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Let $S=\left\{ 1,2,\cdots 2024 \right\}$, if the set of any $n$ pairwise prime numbers in $S$ has at least one prime number, the minimum value of $n$ is \underline{\hspace{2cm}}. | Taking the 15 numbers 1, 22, 32, ..., 432 violates the condition. Furthermore, since $S$ does not contain any non-prime numbers with a minimum prime factor of at least 47, there are only 14 types of non-prime numbers in $S$, excluding 1. Applying the Pigeonhole Principle, we conclude that $n=16$. | Alg1 | 16 |
Let $A_l = (4l+1)(4l+2) \cdots \left(4(5^5+1)l\right)$. Given a positive integer $l$ such that $5^{25l} \mid A_l$ and $5^{25l+1} \nmid A_l$, the minimum value of $l$ satisfying these conditions is \underline{\hspace{2cm}}. | Let $n = 4l$.
Then $A_l=\left(5^5l\right)!{C}_{\left(5^5+1\right)n}^n$, where $\nu_5((5^5l)!) = 5^4n + 5^3n + \cdots + n + \nu_5(n!) = \frac{5^5-1}{4}n + \nu_5(n!)$.
Thus, we need $\nu_{5}\left({C}_{\left(s^{5}+1\right)n}^{n}\right)=\frac{n}{4}-\nu_{5}(n!).$
By Kummer's Theorem, this means in base 5, when adding $n$ and $5^5n$, there are $\frac{n}{4} - \nu_{5}(n!)$ carries.
Notice that $\frac{n}{4} - \nu_5(n!) > \frac{n}{4} - \left(\frac{n}{5}+\frac{n}{25}+\cdots\right) = 0$,
which implies there must be carries when adding $n$ and $5^5n$. Thus, $n$ must have at least 6 digits in base 5.
Suppose $n =a_5a_4\cdots a_0$. The number of carries when adding $n$ and $5^5n$ is the same as the number of carries when adding $a_{5}$ and $\overline{a_{5}a_{4}\cdots a_{0}}$.
Since
\begin{align*}
\frac{n}{4}-v_5(n!) &= \left(\frac{n}{5}+\frac{n}{25}+\cdots\right)-\left(\left[\frac{n}{5}\right]+\left[\frac{n}{25}\right]+\cdots\right) \\
&=\left\{\dfrac{n}{5}\right\}+\left\{\dfrac{n}{25}\right\}+\cdots\\
&=\frac{a_{0}}{5}+\frac{5a_{1}+a_{0}}{25}+\frac{25a_{2}+5a_{1}+a_{0}}{125}+\cdots\\
&=\frac{a_5+a_4+\cdots+a_0}{4}\\
&\geqslant\frac{a_{5}+a_{0}}{4}\geqslant\frac{5}{4}>1
\end{align*}
This indicates that in the quintile system $a_5$ and $\overline{a_5a_4\cdots a_0}$ are carried at least 2 times, then $a_{1}=4$. While,
$$
\frac{a_5+a_4+\cdots+a_0}4\geqslant\frac{a_5+a_1+a_0}4\geqslant\frac{9}{4}>2,
$$
which implies $a_{2}=4$. Continuing this process, we find that $a_1=a_2=\cdots=a_5=4$.
And then $\frac{a_5+a_4+\cdots+a_0}4\in\mathbf{Z}$, we get $a_0=4$.
Obviously, such $n$ is indeed satisfied the requirements. Therefore, the minimum value of $l$ that satisfies the condition is $\frac{5^6-1}4=3906$
Finally, note that the number of carries when adding $5n$ to $5^6n$ is the same as the number of carries when adding $5n$ to $5^{5}n$, which means if $l$ satisfies the conditions, then $5l$ also satisfies the conditions, implying that there are infinitely many values of $l$ satisfying the conditions. | Alg2 | 3906 |
Sasha collects coins and stickers, with fewer coins than stickers, but at least 1 coin. Sasha chooses a positive number $t > 1$ (not necessarily an integer). If he increases the number of coins by a factor of $t$, then he will have a total of 100 items in his collection. If he increases the number of stickers by a factor of $t$, then he will have a total of 101 items in his collection. If Sasha originally had more than 50 stickers, then he originally had \underline{\hspace{2cm}} stickers. | Let $m$ and $n$ be the number of coins and stickers Sasha originally had, respectively.
According to the problem, we have:
\begin{align}
mt+n&=100,\label{Alg3_1} \\
m+nt&=101.\label{Alg3_2}
\end{align}
From \eqref{Alg3_2}$-$\eqref{Alg3_1}, we can learn that
$(n - m)(t - 1) = 1 \Longrightarrow t = 1 + \frac{1}{n - m}$.
From \eqref{Alg3_1}$+$\eqref{Alg3_2}, we can learn that
$(n + m)(t + 1) = 201 \Longrightarrow t = \frac{201}{m + n} - 1$.
Let $a = n - m$ and $b = n + m$.
Since $n > m$, we have $a > 0$.
Comparing the two different expressions in terms of $t$, we get:
$1 + \frac{1}{n - m} = \frac{201}{m + n} - 1 \Leftrightarrow 1 + \frac{1}{a} = \frac{201}{b} - 1 \Leftrightarrow \frac{2a + 1}{a} = \frac{201}{b}$.
Because $2a+ 1$ and $a$ are coprime, $\frac{2n + 1}{a}$ is in lowest terms, which implies that 201 is divisible by $2a+ 1$. Since $201 = 3 \times 67$, it has only four positive divisors: 1, 3, 67, and 201.
Since $2a + 1 > 1$, there are three possible cases:
\begin{enumerate}
\item $2a + 1 = 3$.
Then $a = 1 \Rightarrow \frac{201}{b} = 3 \Rightarrow b = 67$.
Hence, $m = \frac{1}{2}(b - a) = 33$,
$n = \frac{1}{2}(a + b) = 34$,
and $t = 2$,
which does not satisfy the condition.
\item $2a + 1 = 67$.
Then $a = 33 \Rightarrow \frac{201}{b} = \frac{67}{33} \Rightarrow b = 99$.
Hence, $m = \frac{1}{2}(b - a) = 33$,
$n = \frac{1}{2}(a + b) = 66$,
and $t = \frac{34}{33}$.
It can be easily verified that this case satisfies the condition.
\item $2a + 1 = 201$.
Then $a = 100 \Rightarrow \frac{201}{b} = \frac{201}{100} \Rightarrow b = 100$.
Hence, $m = \frac{1}{2}(b - a) = 0$,
and $n = \frac{1}{2}(a + b) = 100$.
Since the number of coins cannot be 0, this case does not satisfy the condition.
\end{enumerate} | Alg3 | 66 |
Let $n$ be a positive integer. An integer $k$ is called a "fan" of $n$ if and only if $0 \leqslant k \leqslant n-1$ and there exist integers $x$, $y$, $z \in \mathbf{Z}$ such that $x^2+y^2+z^2 \equiv 0 (\mathrm{mod} \, n)$ and $xyz \equiv k (\mathrm{mod} \, n)$. Let $f(n)$ denote the number of fans of $n$. Then f(2020) = \underline{\hspace{2cm}}. | For a fan $k$ of 2020, since there exists $x^2+y^2+z^2 \equiv 0 (\mathrm{mod} \, 2020)$, particularly, we have $x^2+y^2+z^2 \equiv 0 (\bmod 4)$.
Thus, $x^2 \equiv 0$ or $1 (\mathrm{mod} \, 4)$, which implies $x$, $y$, $z$ are all even.
Therefore, $k \equiv xyz \equiv 0 (\mathrm{mod} \, 4)$.
Also, $x^2+y^2+z^2 \equiv 0 (\mathrm{mod} \, 5)$, and $x^2 \equiv 0$, $1$, $4 (\mathrm{mod} \, 5)$, so there must be a number among $x$, $y$, $z$ that is a multiple of 5.
Hence, $k \equiv 0 (\mathrm{mod} \, 5)$.
Therefore, a fan $k$ of 2020 must be a multiple of 20.
Next, we prove that all multiples of 20 are fans of 2020.
Since $x^2+y^2+z^2 \equiv 0 (\mathrm{mod} \, 101)$ has solutions, let $x=a$, $y=6a$, $z=8a$, then any $k \equiv xyz \equiv 48a^3 (\mathrm{mod} \, 101)$ is fan of 101.
If there exist $i \neq j$ such that $48i^3 \equiv 48j^3 (\mathrm{mod} \, 101)$ ($0 \leqslant i < j \leqslant 100$), then $(i-j)(i^2+ij+j^2) \equiv 0 (\mathrm{mod} \, 101)$.
Since $i-j \neq 0 (\mathrm{mod} \, 101)$, we have $i^2+ij+j^2 \equiv 0 (\bmod 101)$.
Thus, $(2i+j)^2 \equiv -3j^2 (\mathrm{mod} \, 101)$, implying that -3 is a quadratic residue modulo 101.\label{Alg4_1}
But from the law of quadratic reciprocity, $\left( \frac{3}{101} \right) \left( \frac{101}{3} \right) = (-1)^{\frac{3-1}{2} \cdot \frac{101-1}{2}} = 1$, and $\left( \frac{101}{3} \right) = -1$,
so $\left( \frac{3}{101} \right) = -1$.
Furthermore, $\left( \frac{-3}{101} \right) = \left( \frac{3}{101} \right) \left( \frac{-1}{101} \right)$.
Since $100 \equiv 10^2 \equiv -1 (\mathrm{mod} \, 101)$, we have $\left( \frac{-1}{101} \right) = 1$, which implies $\left( \frac{-3}{101} \right) = -1$, contradicting our assumption\ref{Alg4_1}.
This indicates that when $i$ traverses the complete system modulo 101, $48i^3$ also traverses the complete system modulo 101. Since $0 \leqslant k \leqslant 2019$, we conclude that $f(2020) = 101$. | Alg4 | 101 |
Four positive integers satisfy $a^3=b^2$, $c^5=d^4$, and $c-a=77$. Then, $d-b=$
\underline{\hspace{2cm}}. | Given the conditions, we can assume $a^3=b^2=x^6$ and $c^5=d^4=y^{20}$, which yields $y^4-x^2=(y^2-x)(y^2+x)=77$. Hence, we find that $y= 3, ~x= 2, ~d- b= 243- 8= 235.$ | Alg5 | 235 |
The smallest $n$ such that both $3n+1$ and $5n+1$ are perfect squares is \underline{\hspace{2cm}}. | It can be easily verified. | Alg6 | 16 |
Find the largest positive integer $n$ such that the product of the numbers $n, n+1, n+2,\cdots, n+100$ is divisible by the square of one of these numbers. | When n=100!, $\frac{n(n+1)(n+2)... (n+100)}{n^2}=\binom{n+100}{100}$is an integer.
If $n>100!$, Let the product be divisible by the square of $n+k$, then:
$n(n+1)(n+2)... (n+k-1)(n+k+1)(n+k+2)\cdots (n+100)\equiv0({\mathrm{mod}}n+k)$,
namely, $-1^kk! (100-k)! \equiv0({\mathrm{mod}}n+k)$.
But by $n>100!$, $-1^kk! (100-k)! <n+k$, and $-1^kk! (100-k)!$ non-zero, resulting in contradiction.
So the maximum n is 100! | Alg7 | 100! |
Given a positive integer $x$ with $m$ digits in its decimal representation, and let $x^3$ have $n$ digits. Which of the following options cannot be the value of $m + n$?
\begin{align*}
\text{A)}\ & 2022 &
\text{B)}\ & 2023\\
\text{C)}\ & 2024 &
\text{D)}\ & 2025\\
\end{align*} | Given that $10^{m-1} \leq x < 10^m$,
then $10^{3m-3} \leq x^3 < 10^{3m}$.
Hence, $n$ can take values $3m-2$, $3m-1$, or $3m$.
Thus, $m+n$ cannot be congruent to $1$ modulo $4$.
Therefore, option D is chosen. | Alg8 | D |
Positive integers $a$, $b$, and $c$ satisfy $a > b > c > 1$, and also satisfy $abc \mid (ab - 1)(bc - 1)(ca - 1)$. There are \underline{\hspace{2cm}} possible sets of $(a, b, c)$. | The original statement is equivalent to $abc \mid ab + bc + ca - 1$.
It's evident that $c < 3$, so $c = 2$.
Then, $2ab < ab + 2a + 2b$ implies $b < 4$, so $b = 3$.
Substituting back, we find $a = 5$. | Alg9 | 1 |
There are \underline{\hspace{2cm}} sets of positive integers $a \leq b \leq c$ such that $ab - c$, $bc - a$, and $ca - b$ are all powers of 2. | Only the sets $(2,2,2)$,$(2,2,3)$,$(2,6,11)$ and $(3,5,7)$ satisfy the conditions. | Alg10 | 4 |
Define the function $f(x) =[x[x]]$, where $[x]$ represents the largest integer not exceeding $x$. For example, $[-2.5] = -3$.For a positive integer $n$, let $a_n$ be the number of elements in the range set of $f(x)$ when $x \in [0,n)$. Then the minimum value of $\frac{a_n + 90}{n}$ is \underline{\hspace{2cm}}. | When $x \in [0, 1)$, only one value satisfies $f (x) = 0$ , for the positive integer $k$, when $x \in [k, k + 1)$, $x[x]=kx$,
then $f(x)$ has a total of $k$ values on $[k,k+1)$, and obviously the values on different intervals are not equal to each other, then we have:
$$
a_n=1+1+2+... +(n-1)=\frac{n^2-n+2}{2}
$$
So $\frac{a_n+90}{n}=\frac{\frac{n^2-n+2}{2}+90}{n}=\frac{n}{2} +\frac{91}{n}-\frac{1}{2}$, by the mean value inequality positive integer $n$ should be around $\sqrt{2 \times 91}$. Substituting $n = 13$ and $n = 14$ both yield a result of 13. | Alg11 | 13 |
If a positive integer's sum of all its positive divisors is twice the number itself, then it is called a perfect number. If a positive integer $n$ satisfies both $n-1$ and $\frac{n(n+1)}{2}$ being perfect numbers, then $n=$ \underline{\hspace{2cm}}. | Here we need to use a result from Euler:
$n$ is an even perfect number $\Leftrightarrow$ there exists a prime $p$ such that $2^p - 1$ is prime, and $n = 2^{p-1}(2^p - 1)$.
Now let's use this to solve the problem.
Case 1: $n$ is odd. Then $n-1$ is an even perfect number. We can write $n-1 = 2^{p-1}(2^p - 1)$, where both $p$ and $2^p - 1$ are primes. In this case,
$$
\frac{n\left(n+1\right)}2=\frac12\left(2^{p-1}\left(2^p-1\right)+1\right)\left(2^{p-1}\left(2^p-1\right)+2\right)=\left(2^{p-1}\left(2^p-1\right)+1\right)\left(2^{p-2}\left(2^p-1\right)+1\right).
$$
When $p=2$, $n=7,\quad\frac{n\left(n+1\right)}{2}=28$, in this case, $n-1$ and $\frac{n\left(n+1\right)}{2}$ are both perfect numbers.
When $p\geq3$,let $N=\frac{n\left(n+1\right)}{2}$, then $N$ is odd, and
$$
\frac{n+1}2=4^{p-1}-2^{p-2}+1=\left(3+1\right)^{p-1}-\left(3-1\right)^{p-2}+1,
$$
We know by the binomial theorem that $\equiv3\times(p-1)-(p-2)\times3+1+1+1\equiv6({\mathrm{mod}}9).$
Thus $3 \mid N$, but $3^2 \nmid N $, can set $N = 3k, 3 \times k $, in this case, $\sigma(N)=\sigma(3)$, $\sigma(k)=4\sigma(k)$, but $2N=2({\mathrm{mod}}4)$, so $\sigma(N)\neq2N$, so $\frac{n(n+1)}2$ is not perfect.
Case 2: $n$ is an even number, if $4\mid n$, then $n-1=-1({\mathrm{mod}}4)\Rightarrow n-1$ not a perfect square number, at this point for any $d\mid n-1$, we can know that $d$ and $\frac{n-1}d$ one mod $4$ remains -1, and the other ${\mathrm{mod}}4$ remains 1 from $d\times\frac{n-1}d=n-1\equiv-1({\mathrm{mod}}4)$, leading to $d+\frac{n-1}d\equiv0({\mathrm{mod}}4)$, and $4 \mid \sigma(n-1)$, but $2(n-1)=2({\mathrm{mod}}4)$, so $n-1$ is not perfect.
So, $4 \nmid n$, then, can be set $n=4k+2$, now $N=\frac{n\left(n+1\right)}2=\left(2k+1\right)(4k+3)$ is odd. Because of $\left(2k+1,4k+3\right)=1$, so $\sigma\left(N\right)=\sigma\left(2k+1\right)\sigma\left(4k+3\right).$
Same as above knowable $4 \mid \sigma (4k+3) $, if $\sigma(N)=2N$, we can learn that $4 \mid 2N\Rightarrow 2 \mid N$, this is a contradiction.
To sum up, there is only one $n$ that meets the condition, that is, $n=7.$ | Alg12 | 7 |
Try to find all prime numbers with the shape $p^p+1$($p$ is a natural number) that have no more than 19 digits and what the sum of these prime numbers is. | Obviously $p<19\:.$
If $p$ is odd, then $p^p+1$ is divisible by 2, so $p^p+1=2$ is prime only if $p=1$.
If $p$ has an odd factor, let $p = mk$, of which $m$ is odd, then
$p^{p}+1=p^{\prime}+1=\left(p^{k}\right)^{m}+1$, at this time, $p^k+1 \mid p^{p}+1\:.$
Thus $p^{\rho+1}$is not a prime number.
Thus, $p$ can only be an even number less than 19, with only even factors, i.e. $p=2$,$ 4$, $8$, $16$.
If $p=16$, then
$$
16^{18}=2^{64}=\left(2^{10}\right)^6\cdot16>1000^6\cdot16=16\cdot10^{18},
$$
Then $16 ^{16}+1$is more than 19 digits.
If $p=8$, then
$$
8^8+1=2^{24}+1=\left(2^8\right)^3+1=\left(2^8+1\right)\left(2^{16}-2^8+1\right)\text{is a composite number}.
$$
If $p=4$, then $4^4+1=257$ is prime.
If $p=2$, then $2^2+1=5$ is prime.
So the prime numbers are 2, 5, 257, and their sum is 264. | Alg13 | 264 |
For any positive integer $q_0$, consider a sequence $q_1,q_2,\cdots,q_n$ defined by $q_i=\left(q_{i-1}-1\right)^3+3\left(i=1,2,\cdots,n\right)$. If every $q_i\left(i=1,2,\cdots,n\right)$ is a power of prime, then the maximum possible value of $n$ is \underline{\hspace{2cm}}. | Since $m^3 - m = m(m-1)(m+1) \equiv 0 \pmod{3}$,
we have $q_i = (q_{i-1} - 1)^3 + 3 \equiv (q_{i-1} - 1)^3 \equiv q_{i-1} - 1 \pmod{3}$.
Therefore, among $q_1$, $q_2$, and $q_3$, at least one must be divisible by 3, and this number should be a power of 3.
If $3 \mid (q-1)^3 + 3$, then $3 \mid (q-1)^3$, implying $3 \mid q-1$.
Thus, $3^3 \mid (q-1)^3$.
Since $3 \mid (q-1)^3 + 3$, it follows that $(q-1)^3 + 3$ is a multiple of 3 only when $q_i = 1$, and this occurs only when $i = 0$.
However, when $q_0 = 1$, we get $q_1 = 3$, $q_2 = 11$, and $q_3 = 1003 = 17 \times 59$. Therefore, the maximum value of $n$ is 2. | Alg14 | 2 |
The first digit before the decimal point in the decimal representation of $(\sqrt{2} + \sqrt{5})^{2000}$ is \underline{\hspace{2cm}} and after the decimal point is \underline{\hspace{2cm}}. | $\left(\sqrt2+\sqrt5\right)^{2000}=\left(7+2\sqrt{10}\right)^{1000}.$
Let $a_n=\left(7+2\sqrt{10}\right)^n+\left(7-2\sqrt{10}\right)^n$, then $a_0=2,a_1=14\:.$
Note that $a_n$ is a second-order recursive sequence whose characteristic equation is
$\left[t-\left(7+2\sqrt{10}\right)\right]\left[t-\left(7-2\sqrt{10}\right)\right]=0$ , i.e. $t^2-14t+9=0\:.$
So $a_{n+2}-14a_{n+1}+9a_n=0\:.$
Thus, $a_n$ is a series of integers. Calculate the remainder of the first few terms in the sequence modulo 10:
$$
a_0\equiv2({\mathrm{mod}}10),a_1\equiv4({\mathrm{mod}}10),a_2\equiv8({\mathrm{mod}}10)\:.
$$
$$
a_3\equiv6({\mathrm{mod}}10),a_4\equiv2({\mathrm{mod}}10),a_5\equiv4({\mathrm{mod}}10).
$$
Notice the $a_0\equiv a_4\left({\mathrm{mod}}10\right),a_1\equiv a_5\left({\mathrm{mod}}10\right)$, since $\left\{a_n\right\}$ is second order recursive sequence, then $a_{n+4}\equiv\alpha_n({\mathrm{mod}}10)\:.$
Thus $a_{1000}\equiv a_{996}\equiv a_{992}\equiv\cdots\equiv a_0\equiv2({\mathrm{mod}}10).$
Since $0<7-2\sqrt{10}<1$, then
$0<\left(7-2\sqrt{10}\right)^{1000}<1\:.$
So $\left[\left(7+2\sqrt{10}\right)^{1000}\right]=a_n-1\equiv1\left({\mathrm{mod}}10\right).$
Then, in the decimal representation of $\left(\sqrt{2}+\sqrt{5}\right)^{2000}$, the first digit before the decimal point is 1.
Also because $0<7-2\sqrt{10}<0.9$, then $0<\left(7-2\sqrt{10}\right)^{1000}<0.1\:.$
So
$\left\{7+2\sqrt{10^n}\right\}=1-\left(7-2\sqrt{10}\right)^n>0.9\:.
$
Then, in the decimal representation of $\left(\sqrt{2}+\sqrt{5}\right)^{2000}$, the first digit after the decimal point is 9. | Alg15 | 1,9 |
$N$ is a 5-digit number composed of 5 different non-zero digits, and $N$ is equal to the sum of all three digits formed by 3 different digits in these 5 digits, then the sum of all such 5-digit $N$ is \underline{\hspace{2cm}}. | Let $N=\overline{a_1a_2a_3a_4a_5}$ be the 5-digit number.
There are $P_4^2=12$ three-digit numbers with $a_1$ as the hundreds digit, $P_4^2=12$ three-digit numbers with $a_2$ as the tens digit, and $P_4^2=12$ three-digit numbers with $a_3$ as the units digit. So, according to the given condition:
$$
N=\overline{a_{1}a_{2}a_{3}a_{4}a_{5}}=\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)\left(100\cdot12+10\cdot12+12\right)=1332\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right).
$$
Since $9 \mid 1332$, it follows that $9 \mid N = \overline{a_1a_2a_3a_4a_5}$, and consequently $9\mid (a_{1}+a_{2}+a_{3}+a_{4}+a_{5})$.
Thus, $N$ must be a multiple of $1332 \cdot 9 = 11988$.
Considering the possible values of $a_1+a_2+a_3+a_4+a_5$, we have $15=1+2+3+4+5\leq a_1+a_2+a_3+a_4+a_5\leq9+8+7+6+5=35$. Therefore, $a_1+a_2+a_3+a_4+a_5$ can only be 18 or 27.
\begin{enumerate}
\item When $a_1+a_2+a_3+a_4+a_5=18$,
$\overline{a_{1}a_{2}a_{3}a_{4}a_{5}}=1332\cdot18=23976$,
but $2+3+9+7+6=27\neq18$.
\item When $a_1+a_2+a_3+a_4+a_5=27$,
$\overline{a_{1}a_{2}a_{3}a_{4}a_{5}}=1332\cdot27=35964$,
and $3+5+9+6+4=27$.
Therefore, the required 5-digit number is only 35964.
\end{enumerate} | Alg16 | 35964 |
If the last three digits of a positive integer $n$ cubed are 888, then the minimum value of $n$ is \underline{\hspace{2cm}}. | If the cube of a positive integer ends in 8, then the number itself must end in 2, meaning it can be written in the form $n=10k+2$ (where $k$ is a non-negative integer), and hence $n^3=\left(10k+2\right)^3=1000k^3+600k^2+120k+8$.
The digit in the tens place of $n^3$ is determined by $120k$.
Since we require the tens digit of $n^{3}$ to be 8, the units digit of $12k$ should be 8, meaning the units digit of $k$ must be 4 or 9. Therefore, we can let $k=5m+4$($m$ is a non-negative integer).
Then $n^3=\left\lceil10(5m+4)+2\right\rceil^3=125000m^3+315000m^3+264600m+74088$.
To make the hundreds digit of $n^{3}$ 8, we need the units digit of $2646m$ to be 8. The smallest value of $m$ that satisfies this condition is 3.
Then, $k=5m+4=19$, and $n=10k+2=192$.
It can be verified that $n^3=7077888$, and its last three digits are 888.
Therefore, the minimum value of $n$ is 192. | Alg17 | 192 |
$a $, $b$ are both two-digit positive integers, $100a+b$ and $201a+b$ are both four-digit perfect squares, then $a+b=$ \underline{\hspace{2cm}}. | Let $100a+b=m^2$, $201a+b=n^2$, then
$101a=n^2-m^2=\left(n-m\right)\left(n+m\right),\:m,\:n<100.$
So, $n - m < 100, n + m <200 $, $101\mid \left(m+ n\right).$
Thus, $m+n=101\:.$
By substituting $a=n-m=2n-101$, we obtain
$201(2n-101)+b=n^2$, i.e. $n^2-402n+20301=b\in\left(9,100\right).$
Verify that $n=59,m=101-n=42\:.$
Thus, $a = n-m = 17$, $b = n^2-402n+20301= 64$, i.e.$(a, b) =(17,64)$ .
The answer is 17+64=81. | Alg18 | 81 |
For a positive integer $n$, which can be uniquely expressed as the sum of the squares of 5 or fewer positive integers (where two expressions with different summation orders are considered the same, such as $3^2+4^2$ and $4^2+3^2$ are considered the same expression of 25), then the sum of all the $n$ that satisfy the conditions is \underline{\hspace{2cm}}. | First, prove that for all $n{\geq}17$, there are more than 2 different expressions.
Since every positive integer can be expressed as the sum of the squares of four or less positive integers (Lagrange's four-squares theorem),
there exsit non-negative integer $x_i , y_i, z_i, w_i \left(i = 1, 2, 3, 4 \right) $ that satisfy
$$n-0^2=x_0^2+y_0^2+z_0^2+w_6^2,$$
$$n-1^2=x_1^2+y_1^2+z^2+w_1^2,$$
$$n-2^2=x_2^2+y_2^2+z_2^2+w_2^2,$$
$$n-3^2=x_3^2+y_3^2+z_3^2+w_5^2,$$
$$n-4^2=x_4^2+y_4^2+z_4^2+w_4^2,$$
It follows that
\begin{align*}
n &=x_0^2+y_0^2+z_0^2+w_0^2=1^2+x_1^2+y_1^2+z_1^2+w_1^2=2^2+x_2^2+y_2^2+z_2^2+w_2^2\\
&=3^2+x_3^2+y_3^2+z_3^2+w_3^2=4^2+x_4^2+y_4^2+z_4^2+u_4^2.
\end{align*}
Suppose $n\neq1^2+2^2+3^2+4^2=30$. Then,
\[
\left\{1,2,3,4\right\}\neq\left\{x_0,y_0,z_0,w_0\right\}.
\]
Therefore, there exists $k\in\{1,2,3,4\}\setminus\{x_0,y_0,z_0,u_0\}$, and for such $k$,
$
x_0^2+y_0^2+z_6^2+u_0^2\text{ and }k^2+x_k^2+y_k^2+z_2^2+u_k^2\text{ are distinct}.
$
Because $30=1^2+2^2+3^2+4^2=1^2+2^2+5^2$, it suffices to consider $1\leq n\leq16$.
The following positive integers have two or more different expressions:
\begin{align*}
&4=2^2=1^2+1^2+1^2+1^2 \\
&5=1^2+2^2=1^2+1^2+1^2+1^2+1^2 \\
&8=2^2+2^2=1^2+1^2+1^2+1^2+2^2 \\
&9=3^2=1^2+2^2+2^2 \\
&10=1^2+3^2=1^2+1^2+2^2+2^2 \\
&11=1^2+1^2+3^2=1^2+1^2+1^2+2^2+2^2 \\
&12=1^2+1^2+1^2+3^2=2^2+2^2+2^2 \\
&13=1^2+1^2+1^2+1^2+3^2=1^2+2^2+2^2+2^2 \\
&14=1^2+2^2+3^2=1^2+1^2+2^2+2^2+2^2 \\
&16=4^2=2^2+2^2+2^2+2^2 \\
\end{align*}
However, the following six positive integers have only one unique expression:
\begin{align*}
&1=1^2 \\
&2=1^2+1^2 \\
&3=1^2+1^2+1^2 \\
&6=1^2+1^2+2^2 \\
&7=1^2+1^2+1^2+2^2 \\
&15=1^2+1^2+2^2+3^2 \\
\end{align*}
Therefore, the desired positive integer $n$ is 1, 2, 3, 6, 7, or 15. | Alg19 | 34 |
Given that the product of the digits of a natural number $x$ is equal to $44x-86868$, and the sum of its digits is a perfect cube. Then the sum of all such natural numbers $x$ is \underline{\hspace{2cm}}. | Since $44x \geq 86868$, we have $x \geq \left[ \frac{86868+43}{44}\right] = 1975$.
Thus, $x$ is at least a four-digit number.
On the other hand, if $x$ has $k \geq 5$ digits, then
$ 44x - 86868 > 4 \times 10^k - 10^5 \geq 3 \times 10^k > 9^k,$
which implies $44x - 86868 > p(x)$, where $p(x)$ is the product of the $k$ digits of $x$. This is a contradiction, so $x$ is exactly a four-digit number.
Given that the sum of the digits $S(x)$ satisfies $1 \leq S(x) \leq 36$, we have $S(x) = 1, 8$, or $27$. Obviously, $S(x) = 1$ is not valid.
Since $0 < p(x) \leq 9^4 = 6561$, we have $x \leq \left[\frac{86868+6561}{44} \right]= 2123$.
The only possibilities for $x$ satisfying $1975 \leq x \leq 2123$, $S(x) = 8$ or $27$, and $p(x) \neq 0$ are $1989$, $1998$, $2114$, and $2123$. After checking, we find that only $x=1989$ satisfies the given condition, where the product of its digits equals $44x - 86868$. Therefore, $x=1989$ is the unique solution to this problem.
Hence, the sum of all such natural numbers $x$ is $1989$. | Alg20 | 1989 |
Given \(a\) is a prime number and \(b\) is a positive integer such that \(9(2a+b)^2=509(4a+511b)\), we need to find the values of \(a\) and \(b\). | Since \(9(2a+b)^2=3^2(2a+b)^2\) is a perfect square, it follows that \(509(4a+511b)\) must also be a perfect square. Since \(509\) is a prime number, we can express \(4a+511b\) as {\(509 \times 3^2k^2\)}\label{Alg21_1} .
Hence, the original equation becomes \(9(2a+b)^2=509^2 \times 3^2k^2\), which simplifies to \(2a+b=509k\).
Substituting \(b=509k-2a\) into the equation \ref{Alg21_1}, we get \(4a+511(509k-2a)=509\times 3^2k^2\). Solving this equation yields \(a=\frac{k(511-9k)}{2}\).
Since \(a\) is prime, \(\frac{k(511-9k)}{2}\) must also be prime. Thus, we consider the following cases:
(1) When \(k=1\), \(a=\frac{k(511-9k)}{2}=\frac{511-9}{2}=251\) is prime, and \(b=509k-2a=509-2a=7\).
(2) When \(k=2\), \(a=\frac{k(511-9k)}{2}=\frac{511-18}{2}=493=17\times29\), which is not prime.
(3) When \(k>2\) and \(k\) is odd, \(a=\frac{k(511-9k)}{2}=k\cdot\frac{511-9k}{2}\) is prime. Since \(k>1\), \(\frac{511-9k}{2}=1\), but this equation has no integer solutions.
(4) When \(k>2\) and \(k\) is even, \(a=\frac{k(511-9k)}{2}=\frac{k}{2}(511-9k)\) is prime. Since \(\frac{k}{2}>1\), \(511-9k=1\), but this equation has no integer solutions.
Thus, we conclude that \(a=251\) and \(b=7\). | Alg21 | 251,7 |
Let $\varphi(n)$ denote the number of natural numbers coprime to and less than $n$. Then, when $\varphi(pq)=3p+q$, what is the sum of $p$ and $q$? | (1) Proof: When $p$ and $q$ are distinct primes, $\varphi(pq)=(p-1)(q-1)$.
Since $p$ and $q$ are distinct primes, the natural numbers less than $pq$ and coprime to $pq$ cannot be divisible by $p$ or $q$.
Out of the $pq-1$ natural numbers from $1$ to $pq-1$, $q-1$ numbers are divisible by $p$ and $p-1$ numbers are divisible by $q$. Thus,
$$
\varphi(pq)=pq-1-(p-1)-(q-1)=(p-1)(q-1).
$$
(2) From (1), we have $(p-1)(q-1)=3p+q$, which simplifies to $pq-4p-2q+1=0$, or $(p-2)(q-4)=7$.
$$
\begin{cases}
p-2=1, \\q-4=7,
\end{cases}
$$
$$
\begin{cases}
p-2=7, \\q-4=1,
\end{cases}
$$
$$
\begin{cases}
p-2=-1, \\q-4=-7,
\end{cases}
$$
$$
\begin{cases}
p-2=-7, \\q-4=-1.
\end{cases}
$$
Solving these equations, we get $p=3$, $q=11$.
Hence, the sum of $p$ and $q$ is $14$. | Alg22 | 14 |
Consider the sequence $\{S_n\}$ constructed as follows: $S_1=\{1,1\}$, $S_2=\{1,2,1\}$, $S_3=\{1,3,2,3,1\}$, and in general, if $S_k=\{a_1,a_2,\cdots,a_n\}$, then $S_{k+1}=\{a_1,a_1+a_2,a_2,a_2+a_3,\cdots,a_{n-1}+a_n,a_n\}$. What is the number of terms equal to $1988$ in $S_{1988}$? | Let $\varphi(n)$ denote the Euler's totient function, which counts the number of positive integers less than $n$ that are coprime to $n$.
It's easy to observe that in $S_n$, every pair of adjacent numbers are coprime, and the larger number equals the sum of its two adjacent neighbors.
Now, we use mathematical induction to prove that for $n\geq2$, each pair of coprime numbers $a$ and $b$ less than or equal to $n$ appears adjacent exactly twice in $S_2, S_3, \cdots, S_n$.
By symmetry, it suffices to show that $a$ and $b$ appear adjacent exactly once to the left of $n$. The case $n=2$ is obvious.
Assume the result holds for $n-1$ ($n-1\geq2$). Consider only the left of $2$.
If $a<n$, then $a$ and $b$ are not adjacent in $S_n$.
Otherwise, if $a$ and $b$ are adjacent in $S_n$, then $\alpha-b$ and $b$ are both in $S_{n-1}$ and adjacent. By the inductive hypothesis, $\alpha$ and $b$ are adjacent in $S_2, S_3, \cdots, S_{n-1}$. Thus, $\alpha-b$ and $b$ are adjacent in $S_2, S_3, \cdots, S_{n-2}$. This implies that $a-b$ and $b$ are adjacent in $S_2, S_3, \cdots, S_{n-1}$ at least twice, a contradiction.
Therefore, $a$ and $b$ appear adjacent exactly once to the left of $n$. By symmetry, they appear adjacent exactly once to the right of $n$ as well. Hence, $a$ and $b$ appear adjacent exactly twice in $S_2, S_3, \cdots, S_n$.
Now, consider those first occurrences of $n$ in $S_2, S_3, \cdots, S_n$ (not limited to the left of $2$). These are the numbers in $S_n$ coprime to $n$, less than $n$, and their sum equals $n$. By the above argument, there are exactly $2\varphi(n)$ numbers adjacent to $n$. Thus, the number of occurrences of $n$ in $S_n$ is $\varphi(n)$.
Hence, in $S_{1938}$, the number of occurrences of $1988$ is $\varphi(1988)=\varphi(4)\varphi(7)\varphi(71)=840$. | Alg23 | 840 |
For a natural number $n$, let $S(n)$ denote the sum of its digits. For example, $S(611)=6+1+1=8$. Let $a$, $b$, and $c$ be three-digit numbers such that $a+b+c=2005$, and let $M$ be the maximum value of $S(a)+S(b)+S(c)$. How many sets $(a,b,c)$ satisfy $S(a)+S(b)+S(c)=M$? | Let $a=100a_{3}+10a_{2}+a_{1}$, $b=100b_{3}+10b_{2}+b_{1}$, and $c=100c_3+10c_2+c_1$, where $1\leq a_3,b_3,c_3\leq9$ and $0\leq a_2,b_2,c_2,a_1,b_1,c_1\leq9$.
Define $i=a_1+b_1+c_1$, $j=a_2+b_2+c_2$, and $k=a_3+b_3+c_3$. Given the conditions of the problem, we have $i+10j+100k=2005$, and $i$, $j$, $k$ are each less than or equal to $27$.
We find the possible values of $(i,j,k)$:
$(i,j,k)=(5,0,20),(5,10,19),(5,20,18),(15,9,19),\\
(15,19,18),(25,8,19),(25,18,18)$.
Thus, when $(i,j,k)=(25,18,18)$, $S(a)+S(b)+S(c)=i+j+k$ is maximized.
For $i=25$, the possible pairs $(a_1,b_1,c_1)$ are $(7,9,9)$, $(8,8,9)$, and their permutations, giving $3\times2=6$ possible pairs.
For $j=18$, the possible pairs $(a_2,b_2,c_2)$ are:
$$(0,9,9),(1,8,9),(2,7,9),(2,8,8),(3,6,9),(3,7,8),$$
$$(4,5,9),(4,6,8),(4,7,7),(5,5,8),(5,6,7),(6,6,6)$$
and their permutations,
so there are $6\times7+3\times4+1=55$ possible pairs.
For $k=18$, the possible pairs $(a_3,b_3,c_3)$ are the same as $(a_2,b_2,c_2)$, except $(0,9,9)$ and its permutations are excluded. Therefore, there are $55-3=52$ possible pairs for $(a_3,b_3,c_3)$.
Hence, the number of sets $(a,b,c)$ satisfying the condition is $6\times55\times52=17160$. | Alg24 | 17160 |
For a natural number $n$, let $K(n,0)=\varnothing$. For any non-negative integers $m$ and $n$, define $K(n,m+1)$ as the set of elements $k$ such that $1\leq k\leq n$ and $K(k,m)\cap K(n-k,m)=\varnothing$, then the set $K(2004,2004)$ contains \underline{\hspace{2cm}} elements. | Let's first list out some terms and try to identify a pattern:
$K(1,m)={1}$,$K(2,m)={2}$;
$K(3,m)=\{1,2,3\}$, $K(4,m)=\{4\}$;
$K(5,m)=\{1,4,5\}$, $K(6,m)=\{2,4,6\}$;
$K(7,m)=\{1,2,3,4,5,6,7\}$, $K(8,m)=\{8\}$;
$K(9,m)=\{1,8,9\}$, $K(10,m)=\{2,8,10\}$;
$K(11,m)=\{1,2,3,8,9,10,11\}$;
$K(12,m)=\{4,8,12\}$;
$K(13,m)=\{1,4,5,8,9,12,13\}$;
$K(14,m)=\{2,4,6,8,10,12,14\}$;
$K(15,m)=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$,$K(16,m)=\{16\}$.
It seems $K(n,m)$ only depends on $n$, not on $m$.
Now, let's prove two lemmas using mathematical induction.
\textbf{Lemma 1:} $K(2n,m)=\{2j|j\in K(n,m)\}$.
\textbf{Proof:} Assume the proposition holds for positive integers less than $n$.
For $i=1$, it's clear that $1\in K(2n-1,m)$ and $1\notin K(2n,m)$, then $n\in\mathbb{N}_+$ implies $2j+1\notin K(2n,m)$.
Also, $2j\in K(2n,m) \Longleftrightarrow K(2j,m)\cap K(2n-2j,m)=\varnothing$
$\Longleftrightarrow K(j,m)\cap K(n-j,m)=\varnothing$ (by induction hypothesis)
$\Longleftrightarrow j\in K(n,m)$.
\textbf{Lemma 2:} $K(2^n+i,m)=\{2^n+i\}\cup K(i,m)\cup\{2^n+i-j|j\in K(i,m)\}$, where $1\leq i<2^n$.
\textbf{Proof:} Assume the proposition holds for positive integers less than $2^n+i$.
For any $j\in K(i,m)$, if $j<i$, then
$K(j,m)\cap K(i-j,m)=\varnothing$
$\Rightarrow K(j,m)\cap K(2^n-i,m)=\varnothing$ (by induction hypothesis)
$\Rightarrow j\in K(2^n+i,m)$.
If $j=i$, then $i\in K(i,m)$ and $K(j,m)\cap\{2^n\}=\varnothing$, so
$K(i,m)\cap K(2^n,m)=\varnothing$ and thus $i\in K(2^n+i,m)$.
So, $K(i,m)\subset K(2^n+i,m)$.
For any $j\in K(2^n+i,m)$, if $j<\frac{2^n+i}{2}$, then
$K(j,m)\cap K(2^n+i-j,m)=\varnothing$
$\Rightarrow j\in K(i,m)$ (by induction hypothesis).
If $j=i+k$ and $k<2^{n-1}$, then we need to prove
$K(i+k,m)\cap K(2^n-k,m)\neq\varnothing$.
In binary representation, if $j<\frac{2^n+i}{2}$ and there is no carry when adding $i$ and $k$, then $2^a\in K(i+k,m)\cap K(2^n-k,m)$;
otherwise, if there is a carry at position $t<n$, then $2^t\in K(i+k,m)\cap K(2^n-k,m)$. This concludes the proof of Lemma 2.
Now, let's answer the original question. From Lemma 1 and Lemma 2, we have:
\begin{align*}
|K(2004,m)| &= |K(1002,m)| = |K(501,m)| = |K(256+245,m)| \\
&= 2|K(245,m)|+1 = 2|K(128+117,m)|+1 \\
&= 4|K(117,m)|+3 = 4|K(64+53,m)|+3 \\
&= 8|K(53,m)|+7 = 8|K(32+21,m)|+7 \\
&= 16|K(21,m)|+15 = 16|K(16+5,m)|+15 \\
&= 32|K(5,m)|+31 = 32\times3+31 = 127.
\end{align*}
So, there are $127$ elements in the set $K(2004,2004)$. | Alg25 | 127 |
Let $T=\{0,1,2,3,4,5,6\}$ and $M=\left\{\left.\frac{a_1}{7}+\frac{a_2}{7^2}+\frac{a_3}{7^3}+\frac{a_4}{7^4}\right|a_i\in T,i=1,2,3,4\right\}$. If the elements of $M$ are arranged in descending order, then the 2005th number is \underline{\hspace{2cm}}.
\begin{align*}
\text{A)}\ & \frac{5}{7}+\frac{5}{7^2}+\frac{6}{7^3}+\frac{3}{7^4}&
\text{B)}\ & \frac{5}{7}+\frac{5}{7^2}+\frac{6}{7^3}+\frac{2}{7^4}\\
\text{C)}\ & \frac{1}{7}+\frac{1}{7^2}+\frac{0}{7^3}+\frac{4}{7^4}&
\text{D)}\ & \frac{1}{7}+\frac{1}{7^2}+\frac{0}{7^3}+\frac{3}{7^4} \\
\end{align*} | Let $\left[a_1a_2\cdots a_k\right]_p$ denote a $k$-digit number in base $p$. Multiplying each number in set $M$ by $7^4$, we get:
$$
M'=\left\{a_1\cdot7^3+a_2\cdot7^2+a_3\cdot7+a_4 \mid a_i\in T, i=1,2,3,4\right\}=\left\{\left[a_1a_2a_3a_4\right],\left|a_i\right.\in T, i=1,2,3,4\right\}.
$$
The largest number in $M'$ is $[6666]_7=[2400]_{10}$.
In decimal, the 2005th number in descending order from 2400 is 2400 - 2004 = 396. And $[396]_{10}=[1104]_7$, dividing this number by $7^4$, we obtain the numbers in $M$ as $\frac{1}{7} + \frac{1}{7^2} + \frac{0}{7^3} + \frac{4}{7^4}$. Thus, option C is selected. | Alg26 | C |
Mutually prime positive integers $p_n, q_n$ satisfy $\frac{P_n}{q_n}=1+\frac12+\frac13+\cdots+\frac1n$. The sum of all positive integers $n$ such that $3|p_n$ is \underline{\hspace{2cm}}. | Express $n$ in ternary representation:
$n=\begin{pmatrix}a_ka_{k-1}\cdots a_0\end{pmatrix}_3=a_k\cdot3^k+\cdots+a_1\cdot3^1+a_0\:,$
where $a_j\in\{0,1,2\},j=0,1,2,\cdots,k,\alpha_k\neq0\:.$
Let $A_n$ denote the least common multiple of $1,2,\cdots,n$, then $A_n=3^k\cdot B_n,3\backslash B_n$.
Let $L_n=A_n\cdot\frac{P_n}{q_n}=A_n\left(1+\frac12+\cdots+\frac1n\right)$, then $L_n\in\mathbf{N}_+$, and $3|p_n\Leftrightarrow3^{b+1}|L_n\:.$
Let $S_j= \sum _{1\leq i\leq \frac n{3^j}}\frac 1i, j= 0, 1, 2, \cdots , k$ , then
$$L_n=3^k\cdot B_n\sum_{1\leq i\leq n}\frac{1}{i}=B_n\cdot S_k+3^1\cdot B_n\cdot S_{k-1}+\cdots+3^k\cdot B_n\cdot S_{\text{o}}.(*)
$$
\textbf{Lemma:}When $a_i=0$ or 2, $B_i\cdot S_j\equiv0({\mathrm{mod}}3)$; when $a_i=1$, $B_n\cdot S_j\equiv B_n\left({\mathrm{mod}}3\right)$.
\textbf{Proof:} Since
$
\frac{1}{3m+1}+\frac{1}{3m+2}=\frac{3\left(2m+1\right)}{\left(3m+1\right)\left(3m+2\right)}\text{, we have }B_n\cdot\left(\frac{1}{3m+1}+\frac{1}{3m+2}\right)\equiv0({\mathrm{mod}}3)
$
So when $a_j=0$ or 2, $B_n\cdot Sj\equiv0({\mathrm{mod}}3)$; when $a_j=1$, $B_nS_j\equiv\frac{B_n}{3r+1}\equiv B_n\left({\mathrm{mod}}3\right)$.
Returning to the original question, suppose $3^{k+1}|L_n$. From (*), we have $B_nS_k\equiv0({\mathrm{mod}}3)$.
From the lemma, we know $a_{k}=2, S_{k}=\frac{3}{2}$. If $k=0$, then $n=2$.
When $k\geq1$, from (*), we have
$$
0\equiv B_n\cdot\frac{3}{2}+3^1\cdot B_n\cdot S_{k-1}\big({\mathrm{mod}}9\big)\text{, so }0\equiv B_n\cdot S_{k-1}+B_n\cdot\frac12\equiv B_n\cdot S_{k-1}-B_n\left({\mathrm{mod}}3\right),
$$
thus $B_n\cdot S_{k-1}\equiv B_n\left({\mathrm{mod}}3\right)$. From the lemma, we know $a_{k-1}=1, S_{k-1}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}$.
If $k=1$, then $n=(2,1)_{3}=7$.
When $k{\geq}2$, from (*), we have
$$
\begin{aligned}
&0\equiv B_{n}\cdot\frac{3}{2}+3^{1}\cdot B_{n}\cdot S_{k-1}+3^{2}\cdot B_{n}\cdot S_{k-2}\left({\mathrm{mod}}27\right)\:, \\
&\text{so }0\equiv3\cdot B_{n}\cdot S_{k-2}+B_{n}\cdot\frac{1}{2}+B_{n}\cdot\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}\right) \\
&\equiv3\cdot B_{n}\cdot S_{-2}+B_{n}\cdot\left(2+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}\right) \\
&\equiv3\cdot B_n\cdot S_{k-2}+B_n\left(2-2+2+4\right)\equiv3\cdot\left(B_n\cdot S_{k-2}-B_n\right)\left({\mathrm{mod}}9\right).
\end{aligned}
$$
So $B_n\cdot S_{k-2}\equiv B_n\left({\mathrm{mod}}3\right)$. From the lemma, we know $a_{i-2}=1, S_{i-2}=1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{22}$.
If $k\geq3$, from (*), we have
$$
\begin{aligned}
&0\equiv B_n\cdot\frac{3}{2}+3^1\cdot B_n\cdot\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}\right)+3^2\cdot B_n\cdot S_{k-2}+3^3\cdot B_n\cdot S_{k-3}\left({\mathrm{mod}}81\right), \\
&\text{so }0\equiv B_n\bigg(\left.2+\frac14+\frac15+\frac17\right)+3\cdot B_n\cdot S_{k-2}+3^2\cdot B_n\cdot S_{k-3}\equiv B_n\left(2+7+11+4\right)\\
&+3\cdot B_n\cdot S_{k-2}+3^2\cdot B_n\cdot S_{k-3} \\
&\equiv-3\cdot B_n+3\cdot B_n\cdot S_{k-2}+3^2\cdot B_n\cdot S_{k-3}\left({\mathrm{mod}}27\right), \\
\text{from which }&0\equiv3\cdot B_m\cdot S_{\dot{k}-3}+B_n\left(-1+1+\frac12+\frac14+\cdots+\frac1{22}\right) \\
&\equiv3\cdot B_n\cdot S_{k-3}+B_n\left[-1+\left(1+\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-\frac{1}{2}-1\right)\times2+\left(1+\frac{1}{2}+\frac{1}{4}\right)\right] \\
&\equiv3\cdot B_n\cdot S_{k-3}+B_n\left(5-2\right)\equiv3\cdot B_n\cdot S_{k-3}+3\cdot B_n\left({\mathrm{mod}}9\right).
\end{aligned}
$$
Thus $B_n\cdot S_{k-3}+B_n\equiv0({\mathrm{mod}}3)$, and from the lemma, we know this is impossible. Therefore, the sought-after positive integers $n$ are 2, 7, and 22. | Alg27 | 31 |
Given $x$ and $y$ are prime numbers. The sum of the values of $y$ in the solutions of the indeterminate equation $x^2-y^2=xy^2-19$ is \underline{\hspace{2cm}}. | If $x=y$, then there are obviously no solutions to the equation.
From the given equation, we have $x^y\equiv-19({\mathrm{mod}}y)$. Since $x$ and $y$ are both coefficients and $x\neq y$, then $(x,y)=1$. By Fermat's Little Theorem, we have $x^{y-1}\equiv1({\mathrm{mod}}y)$. Thus, we have $x+19\equiv0\left({\mathrm{mod}}y\right)$.
Similarly, $19-y\equiv0\left({\mathrm{mod}}x\right)$. As $x-y+19\equiv0\left({\mathrm{mod}}y\right)$ and $x-y+19\equiv0\left({\mathrm{mod}}x\right)$, we get $x-y+19\equiv0({\mathrm{mod}}xy)$.
It is evident that $x-y+19\neq0$, thus $x+y+19>|x-y+19|\geq xy$, implying $(x-1)(y-1)<20$. Therefore, $|x-y|<19$ and $x-y+19\geq xy$, i.e., $(x+1)(y-1)\leq18$.
So, when $x\geq5$, we have either $y=2$ or $y=3$. However, $x^2-2^x<0$, $x^3-3^x<0$, and $xy^2-19>0$, leading to contradictions. Hence, $x\leq4$.
It can be verified that the solutions to the original indeterminate equation are (2,3) and (2,7). | Alg28 | 10 |
The number of non-zero integer pairs $(a,b)$ for which $\left(a^3+b\right)\left(a+b^3\right)=(a+b)^4$ holds is \underline{\hspace{2cm}}. | Note that $\left(a^3+b\right)\left(a+b^3\right)=\left(a+b\right)^4$
\begin{align*}
&\Leftrightarrow a^{4}+a^{3}b^{3}+ab+b^{4}=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4} \\
&\Leftrightarrow a^3b^3+2a^2b^2+ab=4a^3b+8a^2b^2+4ab^3 \\
&\Leftrightarrow ab(ab+1)^2=4ab(a+b)^2 \\
&\Leftrightarrow ab\left[(ab+1)^2-4(a+b)^2\right]=0
\end{align*}
Hence, $(a,0)$ and $(0,b)$ are solutions to the given equation, where $a,b\in\mathbf{Z}$.
Additional solutions must satisfy $\left(ab+1\right)^2-4\left(a+b\right)^2=0$.
Since $\left(ab+1\right)^2-4\left(a+b\right)^2=0$, either $ab+1=2(a+b)$ or $ab+1=-2(a+b)$.
We consider two cases:
1. If $ab+1=2(a+b)$, then we have $(a-2)(b-2)=3$.
Thus, we have
\begin{align*}
&\begin{cases}a-2=3,\\b-2=1\end{cases}\text{ or }\begin{cases}a-2=1,\\b-2=3\end{cases}\text{ or }\begin{cases}a-2=-3,\\b-2=-1\end{cases}\text{ or }\begin{cases}a-2=-1,\\b-2=-3.\end{cases} \\
&\text{Solving these, we get} \\
&a=5,b=3;a=3,b=5;a=-1,b=1;a=1,b=-1. \\
&\text{If }ab+1=-2(a+b)\text{, then we have }(a+2)(b+2)=3. \\
&\text{Similarly, solving, we get} \\
&a=1,b=-1;a=-1,b=1;a=-5,b=-3;a=-3,b=-5. \\
&\text{In summary, the set of all possible solutions to the given equation is} \\
&\left\{(a,0)\mid a\in\mathbf{Z}\right\}\bigcup\left\{(0,b)\mid b\in\mathbf{Z}\right\}\cup\left\{(-5,-3),(-3,-5),(-1,1),(1,-1),(3,5),(5,3)\right\}. \\
&\text{Among them, there are 6 pairs of non-zero integer solutions: }(-5,-3),(-3,-5),(-1,1),(1,-1),\\
&(3,5),(5,3).
\end{align*} | Alg29 | 6 |
If the sum of the digits of a natural number $\alpha$ equals 7, then $a$ is called an "auspicious number". Arrange all "auspicious numbers" in ascending order $a_1,a_2,a_3,\cdots$, if $a_n=2005$, then $a_{5n}$=\underline{\hspace{2cm}}. | Because the number of non-negative integer solutions of the equation $x_1+x_2+\cdots+x_t=m$ is $C^m+t-1$, and the number of integer solutions satisfying $x_1\geq1$, $x_i\geq0$ ($i\geq2$) is $C^{m-1-2}_{m-2}$. Now taking $m=7$, we can find that the number of $k$-digit "auspicious numbers" is $P(k)=C^{6}_{k+5}$.
2005 is the smallest "auspicious number" of the form $\overline{2abc}$, and $P(1)=C^6_6=1$, $P(2)=C^7_6=7$, $P(3)=C^8_6=28$. For the four-digit "auspicious numbers" $\overline{1abc}$, the number of such numbers satisfying $a+b+c=6$ is the number of non-negative integer solutions, which is $C^{6}_{6+3-1}=28$.
Because 2005 is the 1st+7th+28th+28th+1st=65th "auspicious number", i.e., $a_{65}=2005$, so $n=65$, $5n=325$.
$P(4)=C^6_9=84$, $P(5)=C^{6}_{10}=210$, and $\sum_{k=1}^6P(k)=330$.
So, the last six five-digit "auspicious numbers" from largest to smallest are: 70000,61000,\\
60100,60010,60001,52000.
Hence, the 325th "auspicious number" is 52000, i.e., $a_{5n}=52000$. | Alg30 | 52000 |
The number of integers $n$ in the interval $1\leq n\leq10^6$ such that the equation $n=x^y$ has non-negative integer solutions $x,y$, and $x\neq n$ is \underline{\hspace{2cm}}. | Let $N\Bigl(x^y\Bigr)$ represent the number of integers $x^y$.
If $1<x^5\leq10^6$, using the principle of inclusion-exclusion, we have
$$
N\left(x^y\right)=N\left(x^2\right)+N\left(x^3\right)+N\left(x^5\right)+N\left(x^7\right)+N\left(x^{11}\right)+N\left(x^{13}\right)+N\left(x^{17}\right)+N\left(x^{19}\right)-N\left(x^6\right)
$$
$$
-N\left(x^{10}\right)-N\left(x^{14}\right)-N\left(x^{15}\right)-N\left(x^{15}\right).
$$
Since there are $10^3-1$ square numbers greater than 1 and less than or equal to $10^6$, we have $N\left(x^{2}\right)=999$.
Similarly, there are $10^2-1$ square numbers greater than 1 and less than or equal to $10^6$, i.e., $N\left(x^{3}\right)=99$.
Since $15^5=819375<10^6$, there are $15-1$ fifth power numbers greater than 1 and less than or equal to $10^6$, i.e., $N\left(x^5\right)=14$.
Continuing this pattern, we can deduce that when $1<x^y\leq10^6$,
$$
N\left(x^y\right)=999+99+14+6+2+1+1-9-2-1-1=1110.
$$
Additionally, when $n=1$, there is a non-negative integer solution with $x>1$ and $y=0$.
Therefore, the number of integers $n$ satisfying the conditions is 1111. | Alg31 | 1111 |
Let $p$ be a real number. If all three roots of the cubic equation $5x^3 - 5(p+1)x^2 + (71p - 1)x + 1 = 66p$ are natural numbers, then the sum of all possible values of $p$ is \underline{\hspace{2cm}}. | \\
\textbf{Approach 1}: Since $5 - 5(p+1) + (77p - 1) + 1 = 66p$, $x=1$ is a natural number solution of the original cubic equation.
By synthetic division, the cubic equation is reduced to the quadratic equation $5x^2 - 5px + 66p - 1=0$(1).
This problem is transformed into finding all real numbers $p$ such that the equation (1) has two natural number solutions.
Let $u$ and $v(u\leq v)$ be the two natural number solutions of equation (1). By Vieta's formulas, we have:
$$
\begin{cases}
v+u=p,(2)\\
vu=\frac{1}{5}(66p-1).(3)
\end{cases}
$$
From (2) and (3), we eliminate $p$ to get $5uv=66(u+v)-1\:.$(4)
From the condition, neither $u$ nor $v$ can be divisible by 2, 3, or 11. So, $u\geq 14$.
Furthermore:
Since 49, $\nu=\frac{66u-1}{5u-66}$.(5)
$u>66/5$ and $u\geq 14$ imply $u\geq 17$.
Since 2, 3, and 11 do not divide $u$, $u\geq 17$.
Also from $\mathcal{v}{\geq}u$, we get $\frac{\text{66}u-1}{5u-66}\geq u$.
Thus, $5u^2-132u+1{\leq}0$, and $u\leq\frac{66+\sqrt{66^2-5}}5<\frac{132}5$.
Since 2, 3, and 11 do not divide $u$, $u$ can only be 17, 19, 23, or 25.
By solving (5) when $u=17, 19, 23, 25$, we find that when $u=17$, $\nu=59$, and both are natural numbers.
\textbf{Approach 2}: From equation (5) in Approach 1, we get
$v=\frac{66u-1}{5u-66}=13+\frac{u+857}{5u-66}=13+\frac15\left(1+\frac{4351}{5u-66}\right)=13+\frac15\left(1+\frac{19\times229}{5u-66}\right)$.
To ensure $v$ is an integer, $5u-66$ must divide 19 or 229.
By considering the prime factors of 19 and 229, we find that $5u-66=19$ when $u=17$, and $5u-66=229$ when $u=59$. Since $5u-66=229$ leads to $v\notin\mathbb{N}$, we discard it.
Therefore, $p=u+v=76$.
\textbf{Approach 3}: From Approach 1, for equation (1) to have natural numbers as solutions, $\Delta=25p^2-4\times5(66p-1)$ must be a perfect square.
Assuming $25p^2-20(66p-1)=q^2$, we get $(5p-132)^2-17404=q^2$. Let $5p-132=m$, then $m^2-q^2=17404$.
This implies both $m$ and $q$ are even. Setting $m=2m_0$ and $q=2q_0$, we get $m_0^2-q_0^2=4351=19\times229$. Solving the system of equations derived from $m>q$, we find $m_0=\pm2176,\pm124$.
Thus, $5p-132=2m_{0}=\pm4352$ or $248$. From equation (5) in Approach 1, $p$ is a natural number. From (3), we conclude that both $u$ and $v$ are odd.
Since (2) implies $p$ is even, we find $p=76$. | Alg32 | 76 |
The number of triples of positive integers $(a, b, c)$ satisfying $a^2 + b^2 + c^2 = 2005$ and $a \leq b \leq c$ is \underline{\hspace{2cm}}. | Since any odd perfect square leaves a remainder of 1 when divided by 4, and any even perfect square is a multiple of 4, it follows that among three squares, there must be two even squares and one odd square.
Let $a = 2m, b = 2n, c = 2k - 1$, where $m, n, k$ are positive integers. The original equation becomes:
\[ m^2 + n^2 + k(k - 1) = 501 (1)\]
Since the remainder when a square is divided by 3 can only be 0 or 1, two cases need to be considered.
(i) If $3 | k(k - 1)$, then both $m$ and $n$ must be multiples of 3. Let $m = 3m_1, n = 3n_1$, and $\frac{k(k - 1)}{3}$ be an integer. This yields:
\[ 3m^2 + 3n^2 + \frac{k(k - 1)}{3} = 167 (2) \]
Solving for $\frac{k(k - 1)}{3} \equiv 167 \equiv 2 \pmod{3}$ gives $\frac{k(k - 1)}{3} = 3r + 2$, and $k(k - 1) = 9r + 6$. (3)Since $k \leq 22$, we can try $k = 3, 7, 12, 16, 21$, which lead to:
\[ \begin{cases} k = 3, \\ m_1^2 + n_1^2 = 55 \end{cases} \]
\[ \begin{cases} k = 7, \\ m_1^2 + n_1^2 = 51 \end{cases} \]
\[ \begin{cases}k = 12, \\ m_1^2 + n_1^2 = 41\end{cases} \]
\[ \begin{cases} k = 16, \\ m_1^2 + n_1^2 = 29 \end{cases} \]
\[ \begin{cases} k = 21,\\ m_1^2 + n_1^2 = 9 \end{cases} \]
Among these, only $k = 12$ and $k = 16$ have positive integer solutions.
For $k = 12$, we get $(m_1, n_1) = (4, 5)$, resulting in $a = 6m_1 = 24, b = 6n_1 = 30, c = 2k - 1 = 23$.
For $k = 16$, we get $(m_1, n_1) = (2, 5)$, resulting in $a = 6m_1 = 12, b = 6n_1 = 30, c = 2k - 1 = 31$.
(ii) If $3 \nmid k(k - 1)$, then either $k$ or $k - 1$ must be divisible by 3. Hence, $k$ is congruent to 2 modulo 3, and $k$ can only be 2, 5, 8, 11, 14, 17, or 20.
For each of these values, we need to check if $\frac{k(k - 1)}{3}$ yields an integer:
For $k = 2$, $m_1^2 + n_1^2 = 499$, which does not have a solution.
For $k = 5$, $m_1^2 + n^2 = 481$, which has solutions $(m, n) = (9, 20)$ or $(15, 16)$.
For $k = 8$, $m_1^2 + n_1^2 = 391$, which does not have a solution.
For $k = 11$, $m^2 + n^2 = 319$, which does not have a solution.
For $k = 14$, $m_1^2 + n_1^2 = 229$, which does not have a solution.
For $k = 17$, $m_1^2 + n_1^2 = 121 = 11^2$, resulting in $(m, n) = (2, 15)$.
For $k = 20$, $m^2 + n_1^2 = 121 = 11^2$, which does not have a solution.
Therefore, there are 7 solutions:
$(23, 24, 30), (12, 30, 31), (9, 18, 40), (9, 30, 32), (4, 15, 42), (15, 22, 36), (4, 30, 33)$. All of these satisfy the original equation. | Alg33 | 7 |
The maximum positive integer $k$ that satisfies $1991^k \mid 1990^{19911992} + 1992^{19911990}$ is \underline{\hspace{2cm}}. | First, let's prove by mathematical induction:
For any odd number $a \geq 3$, for all positive integers $n$, we have $\left(1 + a\right)^{a^\alpha} = 1 + S_n a^{n+1}$, where $S_n$ is an integer and $a \nmid S_n$.
For $n = 1$, we have $\left(1 + a\right)^a = 1 + C_{\mu}^{1}a + C_{\mu}^{2}a^{2} + \cdots + C_{\mu}^{\mu}a^{a} = 1 + a^{2}\left(1 + C_{\mu}^{2} + C_{\mu}^{3}a + \cdots + a^{a-2}\right)$. Since $a$ is odd, $a | C_a^2$, thus $a | C_a^2 + C_a^3\alpha + \cdots + \alpha^{\alpha-2}$, and therefore $a \nmid S_1 = 1 + C_a^2 + \cdots + a^{a-2}$. Hence, equation(1) holds for $n = 1$.
Assume that equation(1) holds for a natural number $n = k_0$. Then
\[ \left(1 + a\right)^{a_0 + 1} = \left[\left(1 + a\right)^{k_0}\right]^a = \left(1 + S_{k_0}a^{k_0+1}\right)^a \]
\[ = 1 + S_{k_{0}}a^{k_{0}+2} + C_{u}^{2}S_{k_{0}}^{2}a^{2k_{0}+2} + \cdots + S_{k_{0}}a^{a(k_{0}+1)} = 1 + S_{k_{0}+1}a^{k_{0}+2}, \]
where $S_{k_0+1} = S_{k_0} + C_{a}^{2}S_{k_0}^{2}a^{k_0} + \cdots + S_{k_0}^{i}a^{a(k_0+1)-k_0-2}$. By the induction hypothesis, $a \nmid S_k$, hence $a \nmid S_{k_0+1}$. Therefore, equation(1) holds for $n = k_0 + 1$.
Hence, equation(1) holds for all natural numbers $n$. Similarly, we can prove:
For any odd number $b \geq 3$, for all positive integers $n$, we have $\left(b-1\right)^{b^n} = -1 + T_n b^{n+1}$, where $T_n$ is an integer and $b \nmid T_n$.
Using(1)and (2), we can find integers $S$ and $T$ such that $1991^\nmid S$, $1991 \nmid T$, and
\[ 1990^{1991^{1992}} + 1992^{1991^{1990}} = T \cdot 1991^{1993} + S \cdot 1991^{1991} = 1991^{1991} \left(T \cdot 1991^2 + S\right). \]
Thus, the maximum $k$ we seek is $1991$. | Alg34 | 1991 |
Let $a$ and $b$ be positive integers such that $79 \mid (a + 77b)$ and $77 \mid (a + 79b)$. Then the smallest possible value of the sum $a + b$ is \underline{\hspace{2cm}}. | Note that
\[ 79 \mid (a + 77b) \Leftrightarrow 79 \mid (a - 2b) \Leftrightarrow 79 \mid (39a - 78b) \Leftrightarrow 79 \mid (39a + b), \]
\[ 77 \mid (a + 79b) \Leftrightarrow 77 \mid (a + 2b) \Leftrightarrow 77 \mid (39a + 78b) \Leftrightarrow 77 \mid (39a + b), \]
so $79 \times 77 \mid (39a + b)$. Thus, $39a + b = 79 \times 77k$, where $k \in \mathbb{N}_+$.
Note that
\[ 39a + 39b = 79 \times 77k + 38b = (78^2 - 1)k + 38b = (78^2 - 39)k + 38(k + b). \]
So, $39 \mid (b + k)$, and we have $b + k \geq 39$. Therefore, $39\alpha + 39b \geq (78^2 - 39) + 38 \times 39$, which implies $a + b \geq 156 - 1 + 38 = 193$.
It is easy to see that $b = 38$ and $\alpha = 155$ satisfy the given conditions. Therefore, $a + b = 193$.
Hence, the minimum value of $(s + n)_{\min}$ = 193. | Alg35 | 193 |
Let $a_i, b_i \ (i=1,2,\ldots,n)$ be rational numbers such that for any real number $x$, we have $x^2 + x + 4 = \sum_{i=1}^{n} (a_{i}x + b_{i})^{2}$. Then the minimum possible value of $n$ is \underline{\hspace{2cm}}. | Since $x^2 + x + 4 = \left(x + \frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2 + 1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2$, it is clear that $n = 5$ is possible. We will now prove that $n = 4$ is not possible.
Proof by contradiction: Suppose $n = 4$. Let $x^2 + x + 4 = \sum_{i=1}^4 (a_ix + b_i)^2$, where $a_i, b_i \in \mathbb{Q}$. Then,
\[ \sum_{i=1}^4 a_i^2 = 1, \quad \sum_{i=1}^4 a_ib_i = \frac{1}{2}, \quad \text{and} \quad \sum_{i=1}^4 b_i^2 = 4. \]
So,
\[ \frac{15}{4}= \left(-\alpha_1b_2 + \alpha_2b_1 - \alpha_3b_4 + \alpha_4b_3\right)^2 + \left(-\alpha_1b_4 + \alpha_3b_1 - \alpha_1b_2 + \alpha_2b_1\right)^2 + \left(-\alpha_1b_4 + \alpha_4b_1 - \alpha_2b_3 + \alpha_3b_2\right)^2. \]
The above expression implies that $a^2 + b^2 + c^2 = 15d^2 \equiv -d^2 \pmod{8}$ has a solution. Without loss of generality, assume that at least one of $a, b, c, d$ is odd and $a^2, b^2, c^2, d^2 \equiv 0, 1, 4 \pmod{8}$. It is clear that the above expression has no solution, leading to a contradiction. Hence, $n = 4$ is not possible. | Alg36 | 5 |
Let $\alpha$ and $\beta$ be the two roots of the equation $x^2 - x - 1 = 0$. Define $\alpha_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$ for $n = 1, 2, \ldots$. For some positive integers $a$ and $b$, with $a < b$, if for any positive integer $n$, $b$ divides $a_n - 2na^n$, then the sum of all such positive integers $b$ is \underline{\hspace{2cm}}. | \begin{enumerate}
\item First, let's prove that for any positive integer $n$, we have $\alpha_{n+2} = \alpha_{n+1} + \alpha_n$. Note that
\[ \alpha^{n+2} - \beta^{n+2} = (\alpha + \beta)(\alpha^{n+1} - \beta^{n+1}) - \alpha\beta(\alpha^n - \beta^n) = (\alpha^{n+1} - \beta^{n+1}) + (\alpha^n - \beta^n), \]
which implies $a_{n+2} = a_{n+1} + a_n$.
\item Given the conditions, we know that $b$ divides $\alpha_1 - 2\alpha$, i.e., $b | 1 - 2\alpha$. Since $b > \alpha$, we have $b = 2\alpha - 1$. Furthermore, for any positive integer $n$, we have $b | \alpha_n - 2n\alpha^n$, $b | \alpha_{n+1} - 2(n+1)\alpha^{n+1}$, and $b | \alpha_{n+2} - 2(n+2)\alpha^{n+2}$. Combining these with $\alpha_{n+2} = \alpha_{n+1} + \alpha_n$ and $b = 2\alpha - 1$ being odd, we get
\[ b | (n+2)a^{n+2} - (n+1)a^{n+1} - na^n. \]
Since $(b, a) = 1$, we have $b | (n+2)a^2 - (n+1)a - n$.
\item By setting $n$ to $n+1$ in the above equation, we get $b | (n+3)a^2 - (n+2)a - (n+1)$. Subtracting this from the previous equation yields $b | a^2 - a - 1$, i.e., $2a - 1 | a^2 - a - 1$. So, $2a - 1 | 2a^2 - 2a - 2$. Since $2\alpha^2 \equiv \alpha \pmod{2\alpha - 1}$, we have $2\alpha - 1 | -\alpha - 2$ and $2\alpha - 1 | -2\alpha - 4$. Thus, $2\alpha - 1 | -5$, implying $2\alpha - 1 = 1$ or $5$. However, $2\alpha - 1 = 1$ leads to $b = \alpha$, which is a contradiction. Hence, $2\alpha - 1 = 5$, giving $\alpha = 3$ and $b = 5$.
\item Now, we need to show that when $a = 3$ and $b = 5$, for any positive integer $n$, we have $5 | (\alpha_n - 2n\alpha^n)$. For $n = 1, 2$, we have $\alpha_1 = 1$ and $\alpha_2 = \alpha + \beta = 1$, so $\alpha_1 - 2 \times 3 = -5$ and $\alpha_2 - 2 \times 2 \times 3^2 = -35$, which confirms the condition. Assuming the condition holds for $n = k, k+1$, we can prove it for $n = k+2$ as well. Hence, $\left(a, b\right) = \left(3, 5\right)$ satisfies the conditions.
\end{enumerate} | Alg37 | 5 |
Let $n$ be a natural number greater than $3$ such that $1+C_n^1+C_n^2+C_n^3$ divides $2^{2000}$. Then, the sum of all such $n$ satisfying this condition is \underline{\hspace{2cm}}. | Since $2$ is a prime number, the problem is equivalent to finding natural numbers $n > 3$ such that
\[ 1+C_n^1+C_n^2+C_n^3 = 2^k \text{ for some } k \in \mathbb{N}, k \leq 2000. \]
We have
\[ 1+C_n^1+C_n^2+C_n^3 = 1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} = \frac{(n+1)(n^2-n+6)}{6}, \]
which means
\[ (n+1)(n^2-n+6) = 3 \times 2^{k+1}. \]
Let's substitute $m = n+1$, then we have
\[ m(m^2-3m+8) = 3 \times 2^{k+1}. \]
Now, let's consider different cases for $m$.
\begin{enumerate}
\item If $m = 2^s$ where $m > 4$ and $s \geq 3$, then $m^{2}-3m+8 = 3 \times 2^{t}$ for some $t \in \mathbb{N}$. If $s \geq 4$, then $m^{2}-3m+8 = 3 \times 2^{t} \equiv 8 \pmod{16}$. So, $t = 3$, which implies $m^{2}-3m+8 = 24$, i.e., $m(m-3) = 16$, which is not possible. Thus, we have only $s = 3$, which gives $m = 8$, i.e., $n = 7$.
\item If $m = 3 \times 2^{u}$ where $m > 4$ and $u \geq 1$, then $m^{2}-3m+8 = 2^{\nu}$ for some $\nu \in \mathbb{N}$. If $u \geq 4$, then $m^{2}-3m+8 = 2^{\nu} \equiv 8 \pmod{16}$. So, $\nu = 3$, which implies $m(m-3) = 0$, which is not possible. Also, for $u = 1$ and $u = 2$, $m^{2}-3m+8$ cannot be a power of $2$. When $m = 3 \times 2^3 = 24$, we find $n = 23$.
\end{enumerate}
Thus, the solutions are $n = 7$ and $n = 23$. Therefore, the sum is $7 + 23 = 30$. | Alg38 | 30 |
In the decimal representation, the product of the digits of $k$ equals $\frac{25}{8}k-211$. Then the sum of all positive integers $k$ satisfying this condition is \underline{\hspace{2cm}}. | Let $k$ be a decimal number, and let $s$ be the product of the digits of $k$.
It's easy to see that $s \in \mathbb{N}$, so $8$ divides $k$ and $\frac{25}{8}k-211 \geq 0$, implying $k \geq \frac{1688}{25}$. Since $k \in \mathbb{N}_+$, we have $k \geq 68$.
Also, since $8$ divides $k$, the units digit of $k$ must be even, making $s$ even as well. Since $211$ is odd, $\frac{25}{8}k$ is odd, implying $16$ divides $k$. Let $k=\overline{a_1a_2\cdots a_i}$, where $0 \leq a_i \leq 9$ for $i=2,3,\cdots,t$ and $1 \leq a_1 \leq 9$. By definition, we have:
\[ S = \prod_{i=1}^t a_i \leq a_1 \times 9^{-i} < a_1 \times 10^{-1} = \overline{a_1}\underbrace{00\cdots0}_{t-1 \text{ digits}} \leq k \:. \]
Therefore, $k > s = \frac{25}{8}k-211$, implying $k \leq 99$.
Since $8$ divides $k$ and $16$ does not divide $k$, we have $k = 72$ or $88$. Upon verification, both $k = 72$ and $k = 88$ satisfy the given condition. Hence, the answer is $72+88=160$. | Alg39 | 160 |
Let $n$ be an integer, and let $p(n)$ denote the product of its digits (in decimal representation). Then the sum of all $n$ such that $10\:p(n)=n^{2}+4n-2005$ is \underline{\hspace{2cm}}. | (1) First, we prove that $p(n)\leq n$.
Assume $n$ has $k+1$ digits, where $k\in\mathbb{N}$. Then $n=10^k\alpha_k+10^{k-1}\alpha_{k-1}+\cdots+10\alpha_1+\alpha_0$, where $a_1,a_2,\cdots,a_k\in\{1,2,\cdots,9\}$. Thus, we have $p(n)=a_9a_1\cdots a_k\leq a_k9^k\leq a_k10^k\leq n$. Therefore, $p(n)\leq n$.
(2) First, note that $n^2+4n-2005\geq0$ implies $n\geq43$.
Furthermore, $n^{2}+4n-2005=10\:p(n)\leq10n$ implies $n\leq47$.
Hence, we deduce that $n\in\{43,44,45,46,47\}$.
Upon checking each case, we find $n=45$. | Alg40 | 45 |
There are some positive integers with more than two digits, such that each pair of adjacent digits forms a perfect square. Then the sum of all positive integers satisfying the above conditions is \underline{\hspace{2cm}}. | It is easy to observe that the perfect squares with two digits are: 16, 25, 36, 49, 64, 81.
Note that, starting from the given digits, there can be at most 1 two-digit perfect square. Therefore, after the first two-digit number is selected, the remaining part of the number is uniquely determined. Since there are no perfect squares starting with 5 or 9, the number cannot start with 25 or 81.
Starting from 16, we get 164 and 1649;
From 36, we get 364 and 3649; From 64, we get 649;
From 81, we get 816, 8164, and 81649.
Therefore, the numbers satisfying the condition are 164, 1649, 364, 3649, 649, 8164, and 81649. | Alg41 | 97104 |
Let $\alpha$ be an integer, and $|\alpha|\leq2005$. The number of values of $\alpha$ that make the system of equations $\begin{cases}x^{2}=y+\alpha,\\y^{2}=x+\alpha\end{cases}$ have integer solutions is \underline{\hspace{2cm}}. | If $(x,y)$ is an integer solution to the given system of equations, subtracting the two equations gives
$$
x^2-y^2=y-x\Longleftrightarrow\left(x-y\right)\left(x+y+1\right)=0.
$$
Consider the following two cases.
$(1)$ When $x-y=0$. Let $x=y=m$ be substituted into the system of equations, resulting in $\alpha=m^{2}-m=m\left(m-1\right)$. It's easy to see that $\alpha$ is the product of two consecutive integers. Thus, $\alpha$ is non-negative, and these numbers do not exceed 2005. Moreover, $45\times44=1980<2005$ and $46\times45=2070>2005$. Since $m$ can take all integers from $1$ to $45$, there are $45$ values of $\alpha$ satisfying this condition.
$(2)$ When $x+y+1=0$. Let $x=m$ and $y=-(m+1)$ be substituted into the system of equations, resulting in $a=m^{2}+m+1=m\left(m+1\right)+1$. It's easy to see that $\alpha$ is one greater than the product of two consecutive integers. Adding 1 to the $\alpha$ obtained in the first case gives the $\alpha$ in the second case. Again, there are $45$ distinct values of $\alpha$ satisfying this condition.
In conclusion, there are a total of $90$ values of $\alpha$ satisfying the condition. | Alg42 | 90 |
Divide the set $S=\{1,2,\cdots,2006\}$ into two disjoint subsets $A$ and $B$ such that:
(1) $B \in A$;
(2) If $a\in A$ and $b\in B$ with $a+b\in S$, then $a+b\in B$;
(3) If $a\in A$, $b\in B$, and $a b\in S$, then $a b\in A$.
The number of elements in set $A$ is \underline{\hspace{2cm}}. | Clearly, $1\in B$ (if not, $1\in A$, and by condition (3), for any $b\in B$, $1\times b=b\in A$, contradiction). For any $a\in A$, by condition (2), $a+1\in B$, thus, for any $k\in\mathbb{N}$, $ka+1\in B$. Hence, $2\in B$ (if not, $2\in A$, and for any $k\in\mathbb{N}$, $2k+1\in B$, leading to $13\in B$, contradiction). Similarly, $3,4,6,$ and $12\in B$, implying that any factor of $\alpha-1$ for $\alpha\in A$ belongs to $B$. By condition (3), for any $a\in A$, we have $2a,3a\in A$. Since $13\in A$, we have $13+1=14\in B$ (if not, $7\in B$, implying $14\in A$, contradiction). Also, $2\times13+1=27\in B$, leading to $9\in B$. Similarly, $3\times13+1=40\in B$, hence $20,10,5\in B$, and $8\in B$ (if not, $8\in A$, implying $8\times5=40\in A$, contradiction). Moreover, $5\times13+1=66\in B$, yielding $33,22,11\in B$. Thus, $\{1,2,\cdots,12\}\subseteq B$, and $13\in A$. By condition (2), for any $k\in\mathbb{N}$ and $i=1,2,\cdots,12$, we have $13k+i\in B$. By condition (3), for any $k\in\mathbb{N}$ and $i=1,2,\cdots,12$, we have $13(13k+i)\in A$, especially $13i\in A$ for $i=1,2,\cdots,12$. If $13^2t\in B$ for some $t\in\mathbb{N}_+$, then by condition (2), $13^2t+13i=13(13t+i)\in B$, contradiction. Therefore, for any $t\in\mathbb{N}_+$, $13^2t\in A$. Thus, $A=\left\{13t\,\middle|\, t=1,2,\cdots,\left\lfloor\frac{2006}{13}\right\rfloor\right\}$ and $B=S-A$. Upon inspection, these sets satisfy the conditions. | Alg43 | 154 |
Let $S$ be a finite set of integers. Suppose that for any two distinct elements $p, q \in S$, there exist three elements $a$, $b$, $c \in S$ (not necessarily distinct, and $a \neq 0$) such that the polynomial $F(x) = \alpha x^2 + bx + c$ satisfies $F(p) = F(q) = 0$. The maximum number of elements in $S$ is \underline{\hspace{2cm}} | It is easy to verify that $S = \{-1,0,1\}$ satisfies the condition. Now, we will prove that $|S|_{\max} = 3$.
(1) At least one of $1$ and $-1$ must belong to $S$. Conversely, assume $a_1, a_2 \in S$ such that by the given condition, there exist $\alpha, b, c \in S$ satisfying $F(a_1) = F(a_2) = 0$. Then, $\frac{c}{a} = a_1a_2 \Rightarrow c = aa_1a_2$. Then, there exists $a_{3} = c \in S$, and repeating this process yields $\alpha_i \ (i=1,2,\cdots) \in S$, but $|\alpha_1| \leq |\alpha_2| < |\alpha_3| < \cdots < |\alpha_k| < \cdots$, which contradicts the fact that $S$ is a finite set.
(2) Without loss of generality, let $1 \in S$. There exists $a_1 \in S (\alpha \neq 1)$. Then, by the given condition, there exist $a, b, c \in S$ such that $a+b+c=0 \Rightarrow b = -a-c$, and $a_{1}+1=-\frac{b}{a}=1+\frac{c}{a}$. Thus, $a_1 = \frac{c}{a} \Rightarrow c = aa_1$.
$(i)$ If $a_1 \geq 2$, then for $\alpha \neq \pm 1$, $|c| > |a_1|$. We can find $\alpha_2 = c \in S (|a_2| > |a_1|)$, leading to $|a_1| < |a_2| < \cdots \in S$, which contradicts the finiteness of $S$. If $a = 1, b = -a_1-1; a = -1, b = a_1+1$, then for any $\alpha = \pm 1$, $|b| > |\alpha_i|$, which also leads to a contradiction.
$(ii)$ If $a_1 \leq -2$, consider $-\frac{b}{a} = a_{1}+1, \frac{c}{a} = a_{1}$. By the assumption, there is no $a \in S$ such that $a \geq 2$. Since $a_1 \leq -2$, $b$ and $c$ have opposite signs. If $a \leq -2$, then $c > |a_1| \geq 2$, which is a contradiction. If $a = -1$, then $b = a_1+1, c = -a_1 \geq 2$, which also contradicts. If $\alpha = 1$, then $b = -a_1-1, c = a_1$. If $a_1 \leq -3$, then $b \geq 2$, which is a contradiction. Thus, we conclude that $a_1 \in \{-2,-1,0\}$.
It is evident that $S = \{-2,-1,0,1\}$ does not satisfy the condition. For example, for $-1, -2 \in S$, $x^2+3x+2=0$, which is impossible. Therefore, $|S|_{\max} = 3$. | Alg44 | 3 |
A natural number whose last four digits are 2022 and is divisible by 2003 has a minimum value of \_\_\_\_\_. | Set this number to be $10000 x+2002$, then
\[
\begin{aligned}
x & =\frac{-2002}{10000}=\frac{1}{10000}=\frac{1+2003 \times 3}{10000} \\
& =\frac{601}{1000}=\frac{661}{100}=\frac{667}{10} \\
& =267(\bmod 2003) .
\end{aligned}\]
So the sought value is 2672002. | Number_Theory45 | 2672002 |
The number of positive integer solution in $\left(x^{2}+2\right)\left(y^{2}+3\right)\left(z^{2}+4\right)=60 x y z $ is \_\_\_\_\_. | First, let's determine the upper bounds for $x, y, z$.
Because,
\[
\begin{aligned}
\left(x^{2}+2\right)\left(y^{2}+3\right) & =x^{2} y^{2}+3 x^{2}+2 y^{2}+6 \\
& >\left(x^{2} y^{2}+4\right)+2\left(x^{2}+y^{2}\right) \\
& \geqslant 4xy+4xy=8xy
\end{aligned}
\]
we have from the original equation
\[
\begin{array}{l}
8xy\left(z^{2}+4\right)<60 xyz, \\
2z^{2}-15z+8<0 .
\end{array}
\]
From (1), it is obvious that $z<8$, and since $z=7$ does not satisfy (1), we have $z \leqslant 6 $.
The right side of the original equation is divisible by 5, and since
\[
\begin{array}{l}
x^{2} \equiv 0, \pm 1(\bmod 5), \\
x^{2}+2 \equiv 1,2,3(\bmod 5), \\
y^{2}+3 \equiv 2,3,4(\bmod 5),
\end{array}
\]
it must be that $z^{2}+4$ is divisible by 5. Thus $z \equiv \pm 1(\bmod 5)$, so$z=1,4,6 $.
If $z=6$, then
$$\left(x^{2}+2\right)\left(y^{2}+3\right)=9xy,$$
but $$\left(x^{2}+2\right)\left(y^{2}+3\right) \geqslant 2 \sqrt{2} x \cdot 2 \sqrt{3} y=4 \sqrt{6} x y>9 xy ,$$ contradiction.\\
If $z=4$, then
$\left(x^{2}+2\right)\left(y^{2}+3\right)=12xy$.
When x=1, $y^{2}+3=4y$, so y=1 or 3.
When x=2, $y^{2}+3=4y$, so y=1 or 3.
Thus $x \geqslant 3$, from (2)
\[
\begin{array}{l}
12y=\left(x+\frac{2}{x}\right)\left(y^{2}+3\right)>x\left(y^{2}+3\right)\geqslant 3\left(y^{2}+3\right),
\end{array}
\]\\
thus, $y^{2}-4y+3<0$, and hence, $y=2$.
Then from (2) we obtain $7\left(x^{2}+2\right)=24 x, 7 \mid x$ ,
so $x \geqslant 7$. But $7\left(x^{2}+2\right)-24 x>x(7 x-24) >0 $, which has no solution.
If $z=1$ , then we again arrive at (2).
Thus, the total number of solutions for this problem is $2 \times 4=8$.
The specific solutions are $(x, y, z)=(1,1,4),(1,3,4),(2,1,4) , (2,3,4),(1,1,1),(1,3,1),(2,1,1),(2,3,1) $. | Number_Theory46 | 8 |
The number of integers satisfying the condition that $x^2+5n+1$ is a perfect square is known to be \_\_\_\_\_. | When $n$ is a positive integer, $(n+1)^{2}<n^{2}+5 n+1<(n+3)^{2}$.
Therefore
\[\begin{array}{l}n^{2}+5n+1=(n+2)^{2}=n^{2}+4 n+4, \\
n=3 .\end{array} \]
When n=0, $n^{2}+5n+1=1$ is a perfect square. \\
When n is a negative integer, let $m=-n$, then m is a positive interger, and $n^{2}+5 n+1=m^{2}-5 m+1$.\\
If $m\leqslant 4 $, then $m^{2}-5m+1<0$ .
If $m=5$ (i.e., $n=-5$ ), then $m^{2}-5m+1=1$ is a perfect square.\\
If $m>5$ , set $k=m-5$ , then $m^{2}-5m+1=k(k+5)+1=k^{2}+5 k+1 $.\\
From the previous results, we know $k=3, m=8, n=-8$ .\\
Therefore, the values of the integer $n$ are $3, 0, -5, -8$. | Number_Theory47 | 4 |
If p, q, r are prime numbers such that $p+q+r=1000$, then the remainder when $p^{2} q^{2} r^{2}$ is divided by 48 is \_\_\_\_\_. | One of $p, q, r$ must be 2. Without loss of generality, let's assume $r=2$. Then p and q are both not 2, and $p+q=1000-2$ .
Because 1000-2-3 is a multiple of 5 and not a prime number, both p and q are not 3.
\[
\begin{array}{l}
p^{2} \equiv q^{2} \equiv 1(\bmod 4), \\
p^{2} \equiv q^{2} \equiv 1(\bmod 3), \\
p^{2} q^{2} \equiv 1(\bmod 12), \\
p^{2} q^{2} r^{2} \equiv 4(\bmod 48),
\end{array}
\] | Alg48 | 48 |
Given $x, y, z$ are integers, and $10x^{3}+20y^{3}+2006 xyz=2007z^{3}$, then the maximum of $x+y+z$ is \_\_\_\_\_. | The left side of the original equation consists of three even terms, so the right side $2007z^{3}$ is also even, implying $z$ is even.
Since $2007z^{3}, 2006xyz$, and $20 y^{3}$ are all divided by 4, $10 x^{3}$ is also divided by 4, making $x$ even.
Further, since $10 x^{3}$,$2006xyz$, and $ 2007 z^{3}$ are all divided by 8, $20 y^{3}$ is also divisible by 8, making y even.
Dividing both sides of the original equation by 8, we get:
\[
10\left(\frac{x}{2}\right)^{3}+20\left(\frac{y}{2}\right)^{3}+2006\left(\frac{x}{2}\right)\left(\frac{y}{2}\right)\left(\frac{z}{2}\right)=2007\left(\frac{z}{2}\right)^{3} .
\]
Similarly, $\frac{x}{2}, \frac{y}{2}, \frac{z}{2}$ are all even,
Therefore, we can replace them with $\frac{x}{4}, \frac{y}{4}, \frac{z}{4}$.
Continuing this process, we find that $\frac{x}{2^{n}}, \frac{y}{2^{n}}, \frac{z}{2^{n}} \quad(where, n=0,1,2, \cdots)$ are all integer solutions of the original equation.
However, when $x \neq 0$, taking $2^{n}>|x|$ implies $\frac{x}{2^{n}}$ is not an integer, hence x=0 .
Similarly, y=0, z=0 .
Therefore, the only integer solution of the original equation is $x=y=z=0$ , leading to $x+y+z=0$. | Alg49 | 0 |
For a positive integer $n$, if the first two digits of $5^{n}$ and $2^{n}$ are the same, denoted as $a$ and $b $ respectively, then the value of the two-digit number $ \overline{ab}$ is \_\_\_\_\_. | Let this two-digit number be $x$,
Then there exist positive integers $k$, $h$ , such that \[\begin{array}{l}10^{k} \cdot x<2^{n}<10^{k}(x+1), \\
10^{n} \cdot x<5^{n}<10^{h}(x+1),\end{array}\]
Multiplying both equations, we get $10^{k+h} x^{2}<10^{n}<10^{k+h}(x+1)^{2} $.
Since x is a two-digit number, $10^{2} \leqslant x^{2}, \quad(x+1)^{2} \leqslant 10^{4}$,
so $10^{k+h+2}<10^{n}<10^{k+h+4}$, which implies n=k+h+3 .
Canceling out $10^{k+h}$ , we get $x^{2}<10^{3}<(x+1)^{2} $.
Since $31^{2}=961,32^{2}=1024$ , we have x=31 , hence $\overline{a b}=31$ . | Alg50 | 31 |
The remainder when $\frac{2020 \times 2019 \times \cdots \times 1977}{44!}$ is divided by 2021 is \_\_\_\_\_. | Firstly, the product of the consecutive 44 integers from 1977 to 2020 is divided by $44!$,
meaning $\frac{2020 \times 2019 \times \cdots \times 1977}{44!}$ is a positive integer. Secondly, 2021 is not a prime number, because $2021=43 \times 47 $.
Since $44 !$ does not have a prime factor of 47 , and $42 \times 47=2021-47=1974$,
thus
\[
\begin{array}{l}
1977=3,1978=4, \cdots, 2020=46, \\
\frac{2020 \times 2019 \times \cdots \times 1977}{44 !}=\frac{46 ! \div 2}{44 !}=\frac{46 \times 45}{2} \\
\equiv \frac{(-1) \times(-2)}{2} \equiv 1 \equiv 1975(\bmod 47) \text {. } \\
\end{array}
\]
Also
\[
\begin{aligned}
\frac{2020 \times 2019 \times \cdots \times 1977}{44 !} & =\frac{2020 \times 2019 \times \cdots \times 1979}{42 !} \times \frac{1978 \times 1977}{43 \times 44} \\
& \equiv \frac{(-1) \times(-2) \times \cdots \times(-42)}{42 !} \times \frac{46 \times 1977}{44} \\
& \equiv \frac{3 \times(-1)}{1}=-3 \equiv 1975(\bmod 43),
\end{aligned}
\]
Thus
$\frac{2020 \times 2019 \times \cdots \times 1977}{44 !} \equiv 1975(\bmod 2021)$ | Number_Theory51 | 1975 |
Given the sequence $\{a_n\}: a_1=1, a_{n+1}=\frac{\sqrt{3}a_n+1}{\sqrt{3}-a_n}$, then
$\sum\limits_{n=1}^{2022}a_n=\_\_\_\_\_$. | It is easy to obtain $a_1=1, a_2=2+\sqrt{3},a_3=-2-\sqrt{3},a_4=-1,a_5=-2+\sqrt{3},a_6=-\sqrt{3},a_7=1$. Then the sequence $\{a_n\}$ is periodic with a period of 6, therefore $\sum\limits_{n=1}^{2022}a_n=337(a_1+a_2+...+a_6)=0.$ | AI-Algebra1 | 0 |
Given \(2bx^2 + ax + 1 - b \geq 0\) holds for \(x \in [-1, 1]\), find the maximum value of \(a + b\). | From the problem statement, we know \(xa + (2x^2 - 1)b \geq -1\) always holds for \(x \in [-1, 1]\). Taking \(x = -\frac{1}{2}\), we get
\[
-\frac{1}{2} (a + b) \geq -1 \Rightarrow a + b \leq 2.
\]
When \(a = \frac{4}{3}\), \(b = \frac{2}{3}\), \(2bx^2 + ax + 1 - b = \frac{3}{4} x^2 + \frac{4}{3} x + \frac{1}{3} = \frac{1}{3} (2x + 1)^2 \geq 0\), which always holds for \(x \in [-1, 1]\). At this time, \(a + b = 2\). Therefore, the maximum value of \(a + b\) is 2. | AI-Algebra3 | 2 |
Given $x,y \in [0,+\infty)$, and satisfying $x^3+y^3+6xy=8$. Then the minimum value of $2x^2+y^2=\_\_\_\_\_$. | According to Euler's formula $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$, it is easy to know $x+y=2$, and from Cauchy's inequality, we know $2x^2+y^2\geq \frac{8}{3}$ | AI-Algebra4 | \frac{8}{3} |
Given $f(x)$ and $g(x)$ are two quadratic functions with the coefficient of the quadratic term being 1 for both. If $g(6)=35,\frac{f(-1)}{g(-1)}=\frac{f(1)}{g(1)}=\frac{21}{20}$, then $f(6)=\_\_\_\_\_$. | Let $f(x)=x^2+ax+b$, $g(x)=x^2+cx+d$.
From the given condition we have:
\begin{align*}
20(1-a+b)=21(1-c+d), & \textcircled{1}\\
20(1+a+b)=21(1+c+d),&\textcircled{2}
\end{align*}
From $\textcircled{1}+\textcircled{2}$, we have $40+40b=42+42d$, then
$20b=1+21d$.//
From $\textcircled{1}-\textcircled{2}$, we have
$-40a=-42c$, then $20a=21c$.
From $g(6)=35$, we have $36+6c+d=35$. So $36+6\times \frac{20}{21}a+\frac{20b-1}{21}=35$. Thus $6a+b=-1$. Then $f(6)=36+6a+b=35$. | AI-Algebra5 | 35 |
Given $(n + 1)^{a+1} - n^{a+1} < n a^(a + 1) < n a^{n+1} - (n - 1)^{a+1}\quad (-1 < a < 0)
\textcircled{1}$. Let$x=\sum\limits_{k=4}^{106}\frac{1}{\sqrt[3]{k}}$, then the integer part of x is \_\_\_\_\_. | In $\textcircled{1}$, take $\alpha=-\frac{1}{3}, n=4,5,\cdots , 10^6$, by adding inequalities, we have ${(10^6+1)}^{\frac{2}{3}}-4^{\frac{2}{3}} < \frac{2}{3}\sum\limits_{k=4}^{10^5}\frac{1}{\sqrt[3]{k}}<{(10^6)}\frac{2}{3}-3^{\frac{2}{3}}$. Then the integer part of $x$ is 14996. | AI-Algebra6 | 14996 |
Let $a_1=\frac{\pi}{6}, a_n \in (0,\frac{\pi}{2})$, and $\tan a_{n+1}\cdot \cos a_n=1 (n\geq 1)$. If $\prod\limits_{k=1}^m \sin a_k=\frac{1}{100}$, then $m=$ \_\_\_\_\_. | From $\tan a_{n+1} \cdot \cos a_n=1
\Rightarrow \tan^2 a_{n+1}-\tan^2 a_n=1 \Rightarrow \tan^2 a_n - \tan^2 a_1 = n-1 \Rightarrow \tan^2 a_n=n-1+\frac{1}{3} \Rightarrow sin a_n=\frac{\sqrt{3n-2}}{\sqrt{3n+1}}$. From $\prod\limits_{k=1}^m \sin a_k=\frac{1}{\sqrt{3m+1}}=\frac{1}{100}$, we have $m=3333$. | AI-Algebra7 | 3333 |
Let $y=f(x)$ be a strictly monotonically increasing function, and let its inverse function be $y=g(x)$. Let
$x_1, x_2$ be the solutions to the equations $f(x)+x=2$ and $g(x)+x=2$ respectively. Then $x_1+x_2=\_\_\_\_\_$. | Given that $f(x)+x$ is strictly monotonically increasing and
$f(x_1)+x_1=2=g(x_2)+x=f(g(x_2))+g(x_2)$.
Therefore, $x_1=g(x_2), x_2=f(x_1)$.
Thus, $x_1+x_2=x_1+f(x_1)=2$. | AI-Algebra8 | 2 |
Let $x_0>0, x_0 \neq \sqrt{3}, Q(x_0,0), P(0,4)$, and the line PQ intersects the hyperbola $x^2-\frac{y^2}{3}=1$ at points A and B. If
$\overrightarrow{PQ}=t\overrightarrow{QA}=(2-t)\overrightarrow{QB}$, then $x_0=\_\_\_\_\_$. | Let $l_{PQ}: y = kx+4(k<0)$, $A(x_1, y_1)$. Then $Q(\frac{4}{-k},0)$. From $\overrightarrow{PQ}=t\overrightarrow{QA} \Rightarrow (-\frac{4}{k},-4)=t(x_1+\frac{4}{k},y_1) \Rightarrow -\frac{4}{k}=t(x_1+\frac{4}{k}), -4=ty_1 \Rightarrow x_1=-\frac{4}{kt}-\frac{4}{k}, y_1=-\frac{4}{t}$.
From point A being on the hyperbola, we get $(48-3k^2)t^2+96t-16k^2+48=0$. Similarly, from $\overrightarrow{PQ}=(2-t)\overrightarrow{QB}$, we can obtain the equation $(48-3k^2)(2-t)^2+96(2-t)-16k^2+48=0 \Rightarrow t+(2-t)=-\frac{96}{48-3k^2} \Rightarrow k=-4\sqrt{2}, x_0=\frac{\sqrt{2}}{2}$. | AI-Algebra9 | \frac{\sqrt{2}}{2} |
Assuming sequence ${F_n}$ satisfying: $F_1=F_2=1, F_{n+1}=F_n+F_{n-1} (n\geq 2)$. Then the number of sets of positive integers $(x,y)$ that satisfy $5F_x-3F_y=1$ is | From the given conditions, we know for any $n \geq 2$, we have $F_{n+1}>F_n$.
Notice that $F_n\in Z_{+}, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, \cdots$.
When $x=1,2$, there does not exist $F_r$ satisfying $5F_x-3F_y=1$.
When $x=3$, in this case, to satisfy $5F_3-3F_y=1$, then $F_y=3$, which means $y=4$.
Thus, $(x,y)=(3,4)$ meets the requirement.
By $5F_x-3F_y=1$, we know $y > x$.
If $x+1$=y, then simplifying \textcircled{1} gets $F_{x-2}-F_{x-3}=1 (x\geq 4)$. Therefore, $x-2=3$ or $4\Rightarrow x=5 or 6$.
Thus, $(x,y)=(5,6)$ or $(6,7)$ meets the requirement.
If $y=x-2$, then $5F_x-3F_y<5F_x-6F_x<0$, it's a contradiction.
Overall, $(x,y)=(3,4)$ or (5,6) or (6,7), a total of 3 sets. | AI-Algebra10 | 3 |
For some positive integers $n$, there exists a positive integer
$k\geq 2$ such that for positive integers $x_1,x_2,...,x_k$ satisfying the given condition, $\sum\limits_{i=1}^{k-1}x_ix_{i+1}=n$, $\sum\limits_{i=1}^{k}x_i=2019$
the number of such positive integers is \_\_\_\_\_. | $n=\sum\limits_{i=1}^{k-1}x_ix_{i+1}(x_1+x_3+\cdots)(x_2+x_4+\cdots)=(x_1+x_3+\cdots)(2019-x_1-x_3-\cdots)$. From $1009\times 1010=1019090 \Rightarrow n=1019090$. When $x_1=1009$, $x_2=1010$, if $k=2$, one can obtain $n=1019090$. Let $x_s$ be the smallest number among all considered values, then equality $n= x_s(\sum\limits_{i=1}^k x_i-x_s)=x_s(2019-x_s)= 2018$ holds true if and only if $x_1=x_2=\cdots=x_{2019}=1$. Let $S=\{x|x\in Z, 1008\times 1011x<1009\times 1010\}$. We will prove the range of $n$ falls in $S$ in the following. By dividing $S$ into 1008 intervals:
$S_{2018}=\{x|x\in Z, 1008\times 1011< x < 1009\times 1010\}$,
$S_i=\{x|x\in Z, i(2019-i) < x <(i+1)(2018-i)\}$, where, $i=1,2,\cdots, 1008$. When $n=1009\times 1010$, the construction is given. If $t\in S_i$ and $t\neq 1009\times 1010$, let $t=(i+1)(2018-i)-a$, $a\in [ 1,2018-2i]$, take $k=4$, $x_1=1$, $x_2=2018-i-a$, $x_3=i$, $x_4=a$, now, $n=(i+1)(2018-i)-a$. Therefore, it proved that every number in set $S$ has corresponding $k$ and $x_i$ meets the problem's criteria. Thus, the sought positive number $n\in [2018, 1019090]$. | AI-Algebra11 | 1017073 |
Considering all non-increasing functions $f:\{1,2,\cdots,10\} \rightarrow \{1,2,\cdots,10\}$, some of these functions have fixed points, while others do not. The difference in the number of these two types of functions is \_\_\_\_\_. | Below, a stronger conclusion is proven: For positive integers $n$, considering all non-increasing functions $f:\{1,2,\cdots,n\} \rightarrow \{1,2,\cdots,n\}$, among these functions, it's demonstrated that the difference in the number of functions with and without fixed points is $C_{2n-2}^{n-1}-C_{2n-2}^{n-2}=\frac{1}{n}C_{2n-2}^{n-1}$.
It's noted that there can be at most one fixed point in function $f$. First, using the method of inserting dividers, the number of non-increasing functions is $C_{n-1+n}^{n-1}=C_{2n-1}^{n-1}$. If a function $f$ has a fixed point, i.e., there exists $c$,
such that $f(c)=c$. When the fixed point is $c$, dividing it into two parts $[1,c-1]$ and $[c+1,n]$ and applying the method of inserting dividers again to calculate the number of non-increasing functions, the number of such functions $f$ with a fixed point is obtained as $C_{n-c+c-1-1}^{c-1}C_{c-1+n-c+1}^{n-c}=(C_{n-1}^{c-1})^2$. Consequently, the total number of functions $f$ with a fixed point is calculated as $\sum\limits_{c=1}^n(C_{n-1}^{c-1})^2=C_{2n-2}^{n-1}$.
As a result, the number of functions $f$ without a fixed point is found as $C_{2n-1}^{n-1}-C_{2n-2}^{n-1}=C_{2n-2}^{n-2}$. Therefore, the sought difference is calculated as $C_{2n-2}^{n-1}-C_{2n-2}^{n-2}=\frac{1}{n}C_{2n-2}^{n-1}$. In this problem, with $n=10$, the answer is 4862. | AI-Algebra12 | 4862 |
Given an integer coefficient polynomial $P(x)$ satisfying:
$P(-1)=-4, P(-3)=-40, P(-5)=-156$. The maximum number of solutions x for $P(P(x))=x^2$ is \_\_\_\_\_. | Notice that, $3|(P(x+3)-P(x))(x \in Z)$. If $x\equiv 0(mod3)$,then
$x^2\equiv P(P(x))\equiv P(P(-3))=P(-40)\equiv P(-1)=-4\equiv -1(mod3)$, contradiction. If $x\equiv 1(mod 3)$, then $x^2 \equiv P(P(x))\equiv P(P(-5))=P(-156)\equiv P(-3)=-40\equiv -1(mod3)$, contradiction. If $x\equiv 2(mod 3)$, then $x^2\equiv P(P(x))\equiv P(P(-1))=P(-4)\equiv P(-1)=-4\equiv -1(mod 3)$, contradiction. So the number of $x$ satisfying $P(P(x))=x^2$ is 0. | AI-Algebra13 | 0 |
Given hyperbola $\Gamma: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes the point $M(3,\sqrt{2})$, line $l$ passes its right focus $F(2,0)$ and cross the right branch of $\Gamma$ at points $A$ and $B$, and cross the y-axis at point P. If $\overrightarrow{PA}=m\overrightarrow{AF}, \overrightarrow{PB}=n\overrightarrow{BF}$, then $m+n=$ \_\_\_\_\_. | From the condition given, it is easy to get the equation of hyperbola $\Gamma$ is $\frac{x^2}{3}-y^2=1$. Let $A(x_1,y_1)$, $B(x_2,y_2)$, $P(0,t)$, from $\overrightarrow{PA}=m\overrightarrow{AF} \Rightarrow
x_1=\frac{2m}{m+1},y_1=\frac{t}{m+1} \Rightarrow
(\frac{2m}{m+1})^2-3(\frac{t}{m+1})^2=3 \Rightarrow m^2-6m-3(t^2+1)=0$.
Similarly, from $\overrightarrow{PB}=n\overrightarrow{BF}$, we get $n^2-6n-3(t^2+1)=0$. Therefore, $m,n(m\neq n)$ are two real roots of the equation $x^2-6x-3(t^2+1)=0$. Thus, $m+n=6$. | AI-Algebra14 | 6 |
Let positive real numbers $x_1, x_2, x_3, x_4$ satisfying $x_1x_2+x_2x_3+x_3x_4+x_4x_1=x_1x_3+x_2x_4$. Then the minimum of $f=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\frac{x_4}{x_1}$ is \_\_\_\_\_. | From $x_1, x_2, x_3, x_4 \in R$, using the mean inequality
$f=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\frac{x_4}{x_1}
\geq 2\sqrt{\frac{x_1x_3}{x_2x_4}} + 2\sqrt{\frac{x_2x_4}{x_1x_3}}
=\frac{2(x_1x_3+x_2x_4)}{\sqrt{x_1x_3x_2x_4}}
=\frac{2(x_1+x_3)(x_2+x_4)}{\sqrt{x_1x_2x_3x_4}}
\geq \frac{8\sqrt{x_1x_3}\sqrt{x_2x_4}}{\sqrt{x_1x_2x_3x_4}}=8$.
The equality holds when $x_1=x_3=1$, $x_2=x_4=2+\sqrt{3}$. | AI-Algebra15 | 8 |
Given sequence $\{a_n\}$ satisfying $a_1=a, a_{n+1}=2(a_n+\frac{1}{a_n})-3$. If $a_{n+1} > a_n (n \in Z_{+})$. The range of real number $a$ is \_\_\_\_\_. | From $a_2-a_1 = \frac{(a-1)(a-2)}{a}>0 \Rightarrow 0<a<2 or a>2.$
(1) When $a>2$, from the induction we can prove $a_n>2
\Rightarrow a_{n+1}-a_n=\frac{(a_n-1)(a_n-2)}{a_n}>0 \Rightarrow a_{n+1}>a_n$. (2) When $0<a<\frac{1}{2}$, $a_2=2(a+\frac{1}{2})-3>2\Rightarrow a_n>2\Rightarrow a_{n+1}>a_n>a_1 (n\geq2)$. When $\frac{1}{2}<1<1$, $a_2=2(a+\frac{1}{a})-3 \in (1,2] \Rightarrow a_n\in (1,2] \Rightarrow a_{n+1}-a_n<0 (n\geq2)$, which does not satisfy the requirement. From the above all, $a\in (0,\frac{1}{2}\cup(2,+\infty))$ | AI-Algebra16 | (0,\frac{1}{2}) \cup (2,+\infty) |
Given positive number $\alpha, \beta, \gamma, \delta $ satisfying $
\alpha+\beta+\gamma+\delta=2\pi$, and $k=\frac{3\tan\alpha}{1+\sec \alpha}=\frac{4\tan \beta}{1+\sec \beta}=\frac{5\tan \gamma}{1+\sec \gamma}=\frac{6\tan\delta}{1+\sec\delta}$, then $k$= \_\_\_\_\_. | From the given condition, we can obtain $k=3\tan\frac{\alpha}{2}=4\tan\frac{\beta}{2}=5\tan\frac{\gamma}{2}=6\tan\frac{\delta}{2}$.
Let $a=\tan\frac{\alpha}{2}$, $b=\tan\frac{\beta}{2}$, $c=\tan\frac{\gamma}{2}$, $d=\tan\frac{\delta}{2}$. Then $0=\tan\frac{\alpha+\beta+\gamma+\delta}{2}=\frac{a+b+c+d-abc-bcd-cda-dab}{1+abcd-ab-ac-ad-bc-bd-cd} \Rightarrow k^3-19k=0 \Rightarrow k=\sqrt{19}$ ($k=0, \sqrt{19}$ is abandoned). | AI-Algebra17 | \sqrt{19} |
Let $A=\{1,2,\cdots,6\}$, function $f:A \rightarrow A$. Mark $p(f)=f(1)\cdots f(6)$. Then the number of functions that make $p(f)|36$ is \_\_\_\_\_. | Because $p(f)|36$, so $p(f)|2^a3^b$, $a, b \in \{0,1,2\}$.
We will count by category in the following. \\
(1) If $b=0$, then the number of choices for $a$ can be $C_6^0(a=0)$, $C_6^1(a=1), C_6^1+C_6^2 (a=2)$, where, $a=2$, the two 2 are in different or the same among $f(1),\cdots,f(6)$.\\
(2) If $b=1$, then there are $C_6^1$ choices for 3. The choices for a can be $C_6^0(a=0),C_6^1(a=1), C_5^1+C_6^2(a=2)$, where, $a=2$, the 2 are among the different or same $f(1),\cdots,f(6)$ but not in the one that contains 3.\\
(3) If $b=2$, then there are $C_6^2$ choices for 3, the choices of $a$ can be $C_6^0(a=0)$, $C_6^1(a=1)$, $C_4^1+C_6^2(a=2)$, where, $a=2$, the two 2 can be in different or same among $f(1),\cdots,f(6)$, but can not be in the two that contains 3.
Therefore, in total, $(C_6^0+2C_6^1+C_6^2)+C_6^1(C_6^0+C_6^1+C_5^1+C_6^2)+C_6^2(C_6^0+C_6^1+C_4^1+C_6^2)=580$ | AI-Algebra18 | 580 |
If unit complex number $a, b$ satisfy $a\bar{b}+\bar{a}b = \sqrt{3}$, then $|a-b|=$ \_\_\_\_\_. | From $|a-b|^2=(a-b)(\bar{a}-\bar{b})=1-a\bar{b}-\bar{a}b+1=2-\sqrt{3} \Rightarrow |a-b|=\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2}$ | AI-Algebra20 | \frac{\sqrt{6}-\sqrt{2}}{2} |
The right focus $F_1$ of the ellipse $\Gamma_1: \frac{x^2}{24}+\frac{y^2}{b^2}=1 (0<b<2\sqrt{6})$ coincides with the focus of the parabola $\Gamma_2: y^2=4px(p \in Z_{+})$. The line $l$ passing through
the point $F_1$ with a positive integer slope
intersects the ellipse $\Gamma_1$ at points A and B,
and intersects the parabola $\Gamma_2$ at points C and D. If $13|AB|=\sqrt{6}|CD|$, then $b^2+p=$ \_\_\_\_\_. | Assume line $l:k(x-p (k\in Z_{+})$, combine $y^2=4px$, we obtain $k^2x^2-2p(k^2+2)x+k^2p^2=0$. Assume $C(x_1,y_1)$, $D(x_2,y_2)$, then $x_1+x_2=-\frac{2p(k^2+2)}{k^2}, x_1x_2=p^2$. So $|CD|^2=\frac{16p^2(1+k^2)^2}{k^4}$. Combine $y=k(x-p)$ and $\frac{x^2}{24}+\frac{y^2}{b^2}=1$, we obtain $(b^2+24k^2)x^2-48pk^2x+24(k^2p^2-b^2)=0$. Assume $A(x_3,y_3)$, $B(x_4,y_4)$, then $x_3+x_4=-\frac{48pk^2}{b^2+24k^2}$, then $x_3x_4=\frac{24(k^2p^2-b^2)}{b^2+24k^2}$, so $|AB|^2=\frac{96b^2(1+k^2)(b^2+24k^2-p^2k^2)}{(b^2+24k^2)^2}$. From $13|AB|=\sqrt{6}|CD|$ and $b^2=24-p^2$, we obtain $13k^2(24-p^2)=p(24-p^2+24k^2)$. From $p^2<24$, p is positive integer, only $p=4$ satisfies the condtion, therefore, $b=2\sqrt{2}$. Thus, $b^2+p=8+4=12$. | AI-Algebra22 | 12 |
Given that $p(x)$ is a quintic polynomial. If
$x=0$ is a triple root of $p(x)+1=0$ and $x=1$ is a triple root of $p(x)-1=0$, then the coefficient of the
$x^3$ term in the expression of $p(x)$ is \_\_\_\_\_. | Let $p(x)+1=x^3(ax^2+bx+c)$, $p(x)-1={(x-1)}^3(lx^2+mx+n)$
Differentiate the above equation to the first and second order, respectively, and set $x=0$, then compare the coefficients of the corresponding equations to solve for the coefficient. We get $l=12, m=6, n=2$. Therefore, $p(x)=12x^5-30x^4+20x^3-1$. | AI-Algebra23 | 20 |
Set $X=\{1,2,\cdots,10\}$, mapping $f:X\rightarrow X$ satisfy:\\
(1) $f\circ f = I_x$, where, $f\circ f$ is a composite mapping, $I_x$ is an identity mapping on X.\\
(2) $|f(i)-i|=2$, for any $i\in X$.\\
Then the number of mapping $f$ is \_\_\_\_\_. | From (1), we know (i) $f(x)=x$, (ii) $f(x)=y (x\neq y)$, $f(y)=x$. Set the number of mapping $f_n$ satisfying the condition when $S_n=\{1,2,\cdots,n\}$. When $f(1)=1$, the number of mapping is $f_{n-1}$.
When $f(1)=2$, then $f(2)=1$, the number of mapping is $f_{n-2}$.
When $f(1)=3$, then $f(3)=1$, if $f(2)=2$, the number of mapping is $f_{n-3}$. If $f(2)=4$, then $f(4)=2$, the number of mapping is $f_{n-4}$. Therefore, $f_n=f_{n-1}+f_{n-2}+f_{n-3}+f_{n-4}$. By calculation, $f_1=1$, $f_2=2$, $f_3=4$, $f_4=8$, $\cdots$, $f_{10}=401$. | AI-Algebra24 | 401 |
Given a integer coefficient polynomial of degree 2022 with leading coefficient is 1, how many roots can it possibly have in the interval (0,1) as maximum? | First, if all 2022 roots of the polynomial are within the interval (0,1), then according to Vieta's formulas, its constant term is the product of these 2022 roots, which also must lie within the interval, thus it cannot be an integer, which is a contradiction. Therefore, the polynomial can have at most 2021 roots in the interval $(0,1)$.
Next, we prove that there exists a leading coefficient 1 integer coefficient polynomial of degree 2022, which has at least 2021 roots in the interval $(0,1)$. Let
$P(x)=x^2022+(1-4042x)(3-4042x)\cdot(5-4042x)\cdots(4041-4042x)$. Note that, for each $k=0,1,\cdots,2021$, we have
$P(\frac{2k}{4042})=(\frac{2k}{4042})^{202}+(-1)^k(2k-1)!!\cdot(4041-2k)!!$.
When k is even, its value is positive; when
k is odd, its value is negative. It is evident that there are at least 2021 sign changes in the interval
$(0,1)$, therefore, it has at least 2021 roots in the interval. | AI-Algebra25 | 2021 |
The system of equations $\left\{\begin{array}{l}x^{2} y+y^{2} z+z^{2}=0, \\ z^{3}+z^{2} y+z y^{3}+x^{2} y=\frac{1}{4}\left(x^{4}+y^{4}\right)\end{array}\right.$ has how many sets of real solution $(x,y,z)$ ?. | \\
First consider the case that $x, y, z$ are all not equal to 0.\\
Set $y=k z$. then the system of equations can be written
$\left\{\begin{array}{l}\frac{x^{2}}{y}+z+\frac{z^{2}}{y^{2}}=0 \\ \frac{4 z^{3}}{y^{2}}+\frac{4 z^{2}}{y}+4 y z+\frac{4 x^{2}}{y}=\frac{x^{4}}{y^{2}}+y^{2}\end{array}\right.$
$\left\{\begin{array}{l}\frac{x^{2}}{y}=-z-\frac{z^{2}}{y^{2}}, \\ \left(\frac{x^{2}}{y}-2\right)^{2}=4 y z+\frac{4 z^{2}}{y}+\frac{4 z^{3}}{y^{2}}-y^{2}+4\end{array} \Rightarrow\left(z+\frac{z^{2}}{y^{2}}+2\right)^{2}=4 y z+\frac{4 z^{2}}{y}+\frac{4 z^{3}}{y^{2}}-y^{2}+4\right.$.\\
put $y=k z$ into the above equation
$\left(z+\frac{1}{k^{2}}+2\right)^{2}=4 k z^{2}+\frac{4 z}{k^{2}}+\frac{4 z}{k}-k^{2} z^{2}+4$
$\Rightarrow\left(4 k-k^{2}-1\right) z^{2}+\left(\frac{2}{k^{2}}+\frac{4}{k}-4\right) z-\left(\frac{1}{k^{4}}+\frac{4}{k^{2}}\right)=0$.\\
Notice that,
$\Delta=\left(\frac{2}{k^{2}}+\frac{4}{k}-4\right)^{2}+4\left(\frac{1}{k^{4}}+\frac{4}{k^{2}}\right)\left(4 k-k^{2}-1\right)$
$=\frac{32}{k^{3}}+\frac{32}{k}-\frac{20}{k^{2}}$.\\
If $k<0$, i.e. $y, z$ have different signs, then, $\Delta<0, z$ has no solution. Thus, the system of equations has no solution.\\
If $k>0$, when $y>0, z>0$, $\frac{x^{2}}{y}+z+\frac{z^{2}}{y^{2}}>0$, is contradiction with (1). Thus, $y<0, z<0$.\\
Solved from the original system of equations
$z^{3}-x^{2} y^{2}+x^{2} y=\frac{1}{4}\left(x^{4}+y^{4}\right)$\\
now, $z^{3}-x^{2} y^{2}+x^{2} y<0, \frac{1}{4}\left(x^{4}+y^{4}\right)>0$, contradiction. Therefore, there must be 0 among $x, y, z$.\\
(1) When $z=0$, the system of equations becomes
$x^{4}+y^{4}=0 \Rightarrow x=y=0$. Thus, $x=y=z=0$.\\
(2) When $y=0$,
$\left\{\begin{array}{l}z^{2}=0 \\ z^{3}=\frac{1}{4} x^{4}\end{array}\right.$ $\Rightarrow x=y=z=0$. \\
(3) When $x=0$,
$\left\{\begin{array}{l}z\left(y^{2}+z\right)=0, \\ z^{3}+z y\left(z+y^{2}\right)=\frac{1}{4} y^{4}\end{array}\right.$ $\Rightarrow z^{3}=\frac{1}{4} y^{4}$. When $y, z$ are not equal to 0 at the same time, then $z>0$.
Therefore, $y^{2}+z=0$, contradiction.Thus, $y=z=0$.\\
Therefore, the only solution of the system of equations is $(x, y, z)=(0,0,0)$. | AI-Algebra26 | 1 |
Set $x, y, z$ are real numbers, satisfying $x^{2}+y^{2}+z^{2}=1$. Then the maximum and minimum of $(x-y)(y-z)(x-z)$ are | Notice that, for any arrangement of $x, y, z$, only change the sign of the required formula, don't change the absolute value of it. Thus, only the maximum needs to be calculated. Set $x \geq y \geq z$. Then, from the mean inequality,
$$
(x-y)(y-z) \leq\left(\frac{(x-y)+(y-z)}{2}\right)^{2}=\left(\frac{x-z}{2}\right)^{2} \Rightarrow(x-y)(y-z)(x-z) \leq \frac{(x-z)^{3}}{4} \text {. }
$$
Thus, only need to prove $x-z \leq \sqrt{2}$. In fact,
$(x-z)^{2}=2 x^{2}+2 z^{2}-(x+z)^{2} \leq 2\left(x^{2}+z^{2}\right)=2-2 y^{2} \leq 2$,
When $x=\frac{1}{\sqrt{2}}, y=0, z=-\frac{1}{\sqrt{2}}$, the equation above holds true. Hence, the maximum value sought is $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$, the minimum value is $-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$. | AI-Algebra28 | \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2} |
If there are a total of 95 numbers, each of which can take any value in +1 or -1, and it is known that the sum of the pairwise products of these 95 numbers is positive, then what is the minimum possible value of this positive sum? | Let the minimum positive value be $\mathrm{N}$, then we have $95+2 \mathrm{~N}=\sum_{i=1}^{95} a_{i}^{2}+2 \mathrm{~N}=\left(\sum_{i=1}^{95} a_{i}\right)^{2}$ being a perfect square, The smallest perfect square larger than 95 is 121, the corresponding $\mathrm{N}$ is 13, constructed with 53 +1s and 42 -1s. | AI-Algebra29 | 13 |
A monotonically increasing sequence of positive integers, starting from the third term, with each subsequent term being the sum of the preceding two terms. If its seventh term is 120, then its eighth term is \_\_\_\_\_. | Set first two terms as $x<y$, Then, from the given conditions, it can be deduced that $5 x+8 y=120$, restricting $x<y$, the only positive integer solution is $x=8, b=10$, hence, it can be further determined that the eighth term is $8 x+13 y=194$. | AI-Algebra30 | 194 |
Sequence $\left\{a_{n}\right\}$ satisfy $a_{0}=0, a_{1}=1$, and for any positive integer $\mathrm{n}$, we have $a_{2 n}=a_{n^{\prime}} a_{2 n+1}=a_{n}+1$, then $a_{2024}=$\_\_\_\_\_. | $a_{2024}=a_{1012}=a_{506}=a_{253}=a_{126}+1=a_{63}+1=a_{31}+2=a_{15}+3=a_{7}+4=$ $a_{3}+5=a_{7}+6=7$, in fact $a_{n}$ is $\mathrm{n}$, which is the number of ones in its binary representations. | AI-Algebra31 | 7 |
Non-negative real numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ satisfy $4 x^{2}+4 y^{2}+z^{2}+2 z=3$, then the minimum of $5 \mathrm{x}+4 \mathrm{y}+3 \mathrm{z}$ is \_\_\_\_\_. | The given conditions, when squared, result in $(2 x)^{2}+(2 y)^{2}+(z+1)^{2}=4$, Combined with the non-negative condition, it can be obtained that $\mathrm{x}, \mathrm{y}, \mathrm{z} \in[0,1]$, Furthermore, by utilizing range scaling, we have $5 x+4 y+3 z \geq 4 x^{2}+4 y^{2}+3 z \geq 4 x^{2}+4 y^{2}+z^{2}+2 z=3$. | AI-Algebra32 | 3 |
Real numbers $x, y$ satisfy $2 x-5 y \leq-6$ and $3 x+6 y \leq 25$, then the maximum value of $9 x+y$ is \_\_\_\_\_. | By adjusting coefficients, we can obtain $27(9 x+y)=47(3 x+6 y)+51(2 x-5 y) \leq 47 \times 25+51 \times(-6)=869$. | AI-Algebra33 | \frac{869}{27} |
The sum of the maximum element and minimum element in set $\left\{\left.\frac{3}{a}+b \right\rvert\, 1 \leq a \leq b \leq 2\right\}$ is \_\_\_\_\_. | The maximum element needs to make $a$ as small as possible, $b$ as large as possible, therefore it is obvious $3+2=5$, while the minimum element needs to make $a$ as large as possible, $\mathrm{b}$ as small as possible, therefore, we have $a=\mathrm{b}$, and utilizing the mean inequality we select $a=\mathrm{b}=\sqrt{3}$, the the minimum element is $2 \sqrt{3}$. | AI-Algebra34 | 5+2 \sqrt{3} |
A monotonically increasing function $\mathrm{f}(\mathrm{x})$ defined on $R^{+}$ satisfies $\mathrm{f}\left(\mathrm{f}(\mathrm{x})+\frac{2}{\mathrm{x}}\right)=-1$ consistently holds within its domain, then $\mathrm{f}(1)=$\_\_\_\_\_. | According to the given conditions, it can only be $\mathrm{f}(\mathrm{x})+\frac{2}{x}$ is a constant value, set this constant to be $\mathrm{t}>0$, then we have $\mathrm{f}(\mathrm{t})+\frac{2}{t}=t$ and $\mathrm{f}(\mathrm{t})=-1$, solved as $\mathrm{t}=1$ (neglecting the negative root), therefore $\mathrm{f}(1)=-1$. | AI-Algebra35 | -1 |
Given set: $A=\left\{x+y \left\lvert\, \frac{x^{2}}{9}+y^{2}=1\right., x+y \in \mathbf{Z}_{+}\right\}, B=\{2 x+y \mid x, y \in A, x<y\}$, $C=\{2 x+y \mid x, y \in A, x>y\}$. Then the sum of all elements in $B \cap C$ is | Set $x=3 \cos \theta, y=\sin \theta$. Then $x+y=\sqrt{10} \sin (\theta+\varphi)$. Thus, $A=\{1,2,3\}$. Therefore, through enumeration we obtain $B=\{4,5,7\}, C=\{5,7,8\} \Rightarrow B \cap C=\{5,7\}$. So, $5+7=12$. | AI-Algebra37 | 12 |
Given a hyperbola $\Gamma: \frac{x^{2}}{7}-\frac{y^{2}}{5}=1$, a line $l: a x+b y+1=0$ intersects $\Gamma$ at point $A$. A tangent to $\Gamma$ drawn through point $A$ is perpendicular to the line $l$. Then $\frac{7}{a^{2}}-\frac{5}{b^{2}}=$\_\_\_\_\_. | Set point $A\left(x_{0}, y_{0}\right)$. Then the tangent is $\frac{x_{0} x}{7}-\frac{y_{0} y}{5}=1$, slope is $\frac{5 x_{0}}{7 y_{0}}$. Additionally, the slope of $l$ is $-\frac{a}{b}$, then from the given condition $\frac{5 a x_{0}}{7 b} y_{0}=1 \Rightarrow x_{0}=\frac{7 b y_{0}}{5 a}$. Combining point $A$ is on line $l$, we have $x_{0}=-\frac{b}{a} y_{0}-\frac{1}{a}$. Therefore, $y_{0}=-\frac{5}{12 b}, x_{0}=-\frac{7}{12 a}$.
Substituting into the equation of the hyperbola $\Gamma$, we obtain $\frac{7}{144 a^{2}}-\frac{5}{144 b^{2}}=1 \Rightarrow \frac{7}{a^{2}}-\frac{5}{b^{2}}=144$. | AI-Algebra38 | 144 |
In the Cartesian coordinate plane $x O y$, point $A(a, 0), B(0, b), C(0,4)$, moving point $D$ satisfies $|C D|=1$. If the maximum value of $|\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O D}|$ is 6, then the minimum value of $a^{2}+b^{2}$ is \_\_\_\_\_. | From moving point $D$ satisfying $|C D|=1$, we know point $D$ lies on a circle with center $C$ and radius 1.
Set $D(x, y)$, then $|\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O D}|=\sqrt{(x+a)^{2}+(y+b)^{2}}$, i.e., the maximum value of distance between point $(x, y)$ and point $(-a,-b)$ is 6, this indicates that point $(-a,-b)$ lies on a circle with center $C$ and radius 5. So $a^{2}+b^{2}=(-a)^{2}+(-b)^{2} \geq 1$, when $a=0, b=1$, the equation holds true. | AI-Algebra39 | 1 |
Given $n$ is positive integer, for $i=1,2, \cdots, n$, positive integer $a_{i}$ and positive even number $b_{i}$ satisfy $0<\frac{a_{i}}{b_{i}}<1$, and for any positive integers $i_{1}, i_{2}\left(1 \leq i_{1}<i_{2} \leq n\right), a_{i_{1}} \neq a_{i_{2}}$ and $b_{i_{1}} \neq b_{i_{2}}$ at least one holds true. If for any positive integer $n$ and all positive integers $a_{i}$ and positive even numbers $b_{i}$ satisfy the above conditions, we all have $\frac{\sum_{i=1}^{n} b_{i}}{n^{\frac{3}{2}}} \geq c$, then the maximum value of real number $c$ is \_\_\_\_\_. | For any positive integer $t$, when $n=t^{2}$, take integers that satisfy conditions:
$a_{1}=1, b_{1}=2 ; a_{2}=1, b_{2}=4 ;$
$a_{3}=2, b_{3}=4 ; a_{4}=3, b_{4}=4 ;$
$\cdots$
$a_{(t-1)^{2}+1}=1, b_{(t-1)^{2}+1}=2 t ;$
$a_{(t-1)^{2}+2}=2, b_{(t-1)^{2}+2}=2 t ;$
$a_{t^{2}}=2 t-1, b_{t^{2}}=2 t$.\\
Then $\sum_{i=1}^{n} b_{i}=\sum_{j=1}^{t}(2 j-1) 2 j=4 \sum_{j=1}^{t} j^{2}-2 \sum_{j=1}^{t} j=\frac{2 t(t+1)(2 t+1)}{3}-t(t+1)=\frac{t(t+1)(4 t-1)}{3}$.\\
So, $c \leq \frac{\sum_{i=1}^{n} b_{i}}{n^{\frac{1}{2}}}=\frac{t(t+1)(4 t-1)}{3\left(t^{2}\right)^{\frac{3}{2}}}=\frac{(t+1)(4 t-1)}{3 t^{2}} \rightarrow \frac{4}{3}(t \rightarrow+\infty)$.\\
Therefore, $c \leq \frac{4}{3}$.\\
The following is proved by mathematical induction: for each positive integer $n$ and all positive integers $a_{i}$ satisfying the conditions and all positive even numbers
$b_{i}(i=1,2, \cdots, n)$, we all have $\frac{\sum_{i=1}^{n} b_{i}}{n^{\frac{3}{2}}} \geq \frac{4}{3}$.\\
When $n=1$, since $b_{1} \geq 2$, then $\frac{b_{1}}{1^{\frac{3}{2}}} \geq 2>\frac{4}{3}$.\\
Assume that equation (1) holds true, when $n=k$ ($k$ is positive integer). When $n=k+1$, consider positive integers $a_{i}$ and positive even numbers $b_{i}(i=1,2, \cdots, k+1)$ satisfying conditions. Next, we aim to prove that there exists $i_{0} \in\{1,2, \cdots, k+1\}$, such that $b_{i_{0}} \geq 2 \sqrt{k+1}$.\\
Assume that the conclusion is not true, then for each $i=1,2, \cdots, k+1$, we have $2 \leq b_{i}<2 \sqrt{k+1}$. Without loss of generality, suppose that the sequence $\left\{b_{i}\right\}(i=1,2, \cdots, k+1)$ is arranged in increasing order. Set $\sqrt{k+1}=m+a$, where, $m=[\sqrt{k+1}]$ ( $[x]$ denotes the greatest integer not exceeding $x$), $0 \leq a<1$. Then $2 \leq b_{i}<2 \sqrt{k+1}=2 m+2 a<2 m+2 \Rightarrow 2 \leq b_{i} \leq 2 m+1$. Since $b_{i}$ is even number, thus, $2 \leq b_{i} \leq 2 m$.\\
For positive integer $j(1 \leq j \leq m)$, the number of positive integers $a_{i}$ satifying $b_{i}=2 j$ is at most $2 j-1$, i.e., $a_{i} \in\{1,2, \cdots, 2 j-1\}$. Therefore, $n=k+1 \leq \sum_{j=1}^{m}(2 j-1)=m^{2}$.\\
Thus, $m \geq \sqrt{k+1} \geq m$, which implies that all equalities hold, i.e., $m=\sqrt{k+1}$, $n=k+1=\sum_{j=1}^{m}(2 j-1)=m^{2}$, $b_{k+1}=2 \sqrt{k+1}$, contradiction.
Therefore, there exists $i_{0} \in\{1,2, \cdots, k+1\}$, such that $b_{i_{0}} \geq 2 \sqrt{k+1}$.\\
From the induction hypothesis, we can obtain
$\sum_{i=1}^{k+1} b_{i}=\sum_{\substack{i=1 \\ i \neq i_{0}}}^{k+1} b_{i}+b_{i_{0}}>\frac{4}{3} k^{\frac{3}{2}}+2 \sqrt{k+1}$.\\
To prove that equation (1) holds when n=k+1$ it suffices to show \frac{4}{3} k^{\frac{3}{2}}+2 \sqrt{k+1} \geq \frac{4}{3}(k+1)^{\frac{2}{2}}$. Notice that, equation (2)
$\Leftrightarrow \frac{4}{3} k \sqrt{k}+2 \sqrt{k+1} \geq \frac{4}{3}(k+1) \sqrt{k+1}$
$\Leftrightarrow 2 k \sqrt{k}+3 \sqrt{k+1} \geq 2(k+1) \sqrt{k+1}$
$\Leftrightarrow \sqrt{k+1} \geq 2 k(\sqrt{k+1}-\sqrt{k})$.\\
From $2 k(\sqrt{k+1}-\sqrt{k})=\frac{2 k}{\sqrt{k+1}+\sqrt{k}}<\frac{k}{\sqrt{k}}=\sqrt{k}<\sqrt{k+1}$.\\
Then equation (3) holds, thus equation (2) holds, thereby completing the proof of equation (1). From equation (1), we know that when $c=\frac{4}{3}$, the original inequality holds.\\
Therefore, $c_{\text {max }}=\frac{4}{3}$. | AI-Algebra40 | \frac{4}{3} |
In a cube $ABCD-A_1B_1C_1D_1$, $AA_1=1$, E, F are the midpoints of edges $CC_1, DD_1$, then the area of the cross-section obtained by the plane AEF intersecting the circumscribed sphere of the cube is \_\_\_\_\_. | Taking $A$ as the origin, and $AB, AD, AA_1$ as the $x,y,z$ axes to establish a spatial rectangular coordinate system, then, $A(0,0,0)$,$E(1,1,\frac{1}{2})$,$F(0,1,\frac{1}{2})$, so $\overrightarrow{AE}=(1,1,\frac{1}{2}), \overrightarrow{AF}=(0,1,\frac{1}{2})$. Let the normal vector of plane $AEF$ be $n=(x,y,z)$. Then,
\[
\left\{
\begin{aligned}
\mathbf{n} \cdot \mathbf{AE} &= 0 \\
\mathbf{n} \cdot \mathbf{AF} &= 0
\end{aligned}
\right
\Rightarrow
\left\{
\begin{aligned}
x + y + \frac{1}{2}z &= 0, \\
y + \frac{1}{2}z &= 0.
\end{aligned}
\right.
\]
Take $n=(0,-1,2)$, the distance from the center of the sphere $O(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ to the plane $AEF$ is $d=\frac{|\overrightarrow{AO}|\cdot n}{|n|} = \frac{\sqrt{5}}{10}$. Let the radius of the cross-sectional circle be $r$, because the radius of the circumscribed sphere of the cube is $R=\frac{\sqrt{3}}{2}$, therefore,$r^2=R^2-d^2=\frac{3}{4}-\frac{1}{20}=\frac{7}{10}$, thus the area of the cross-section is $\frac{7}{10}\pi$. | AlChallenge_Geo1 | \frac{7}{10}\pi |
In tetrahedron $ABCD$, triangle $ABC$ is an equilateral triangle, $\angle BCD=90^{\circ}$, $BC=CD=1$,$AC=\sqrt{3}$,$E$ and $F$ are the midpoints of edges $BD$ and $AC$ respectively. Then the cosine of the angle formed by lines $AE$ and $BF$ is \_\_\_\_\_. | Take the midpoint M of CE, then $\angle MFB$ is the angle formed by AE and BF, where $FM=\frac{1}{2}AE=\frac{\sqrt{6}}{4}$. Hence, $cos\angle MFB=\frac{\frac{3}{16}+\frac{3}{4}-\frac{10}{16}}{2\times\frac{\sqrt{6}}{4}\times\frac{\sqrt{3}}{2}}=\frac{\sqrt{2}}{3}$. | AIChallenge_Geo2 | \frac{\sqrt{2}}{3} |
Let P be a point inside triangle ABC, and 2$\overrightarrow{PA} = \overrightarrow{PB} + \overrightarrow{PC}=0$. If $\angle BAC = \frac{\pi}{3}, BC=2$, then the maximum value of $\overrightarrow{PB}\cdot\overrightarrow{PC}$ is \_\_\_\_\_. | Let M be the midpoint of BC. Then $2\overrightarrow{PB}=\overrightarrow{PB}+\overrightarrow{PC}=-2\overrightarrow{PA}$, hence, P is the midpoint of the median AM of edge BC.
Therefore, $\overrightarrow{PB}\cdot\overrightarrow{PC}=(\overrightarrow{PM}+\overrightarrow{MB})\cdot(\overrightarrow{PM}+\overrightarrow{MC})=|\overrightarrow{PM}|^2+\overrightarrow{MB}\cdot\overrightarrow{MC}=|\overrightarrow{PM}|^2-1=\frac{1}{4}|\overrightarrow{AM}|^2-1$.
Also, since $\angle BAC =\frac{\pi}{3}$, and $BC = 2$, the locus of point A is a circle, where BC is a chord on the circle, and the inscribed angle corresponding to BC is $60^{\circ}$, then $AM \leq \sqrt{3}$.
Therefore, $\overrightarrow{PB}\cdot\overrightarrow{PC}=\frac{1}{4}|\overrightarrow{AM}|^2-1<leq\frac{3}{4}-1=-\frac{1}{4}$. | AIChallenge_Geo3 | -\frac{1}{4} |
A rectangular solid whose length, width, and height are all natural numbers, and the sum of all its edge lengths equals its volume, is called a "perfect rectangular solid. The maximum value of the volume of a perfect rectangular solid is \_\_\_\_\_. | Let the length, width, and height be $a, b, c$, and $a > b > c > 1$. Then $4(a+b+c)=abc \Rightarrow a=\frac{4(b+c)}{b c-4} \Rightarrow b c>4 \Rightarrow b^{2} > b c>4$. Since $a>b$, we have\\
$8 b > \left(b^{2}-4\right) c > b^{2}-4 \Rightarrow\left\{\begin{array}{l}b \geq 3, \\ (b-4)^{2} > 20\end{array} \Rightarrow 3 < b < 8\right.$.\\
Thus $(a, b, c)=(10,3,2),(6,4,2),(24,5,1),(14,6,1),(9,8,1)$.
Therefore, the sought maximum is $24 \times 5 \times 1=120$. | AIChallenge_Geo4 | 120 |
In the convex quadrilateral $A B C D$ inscribed in a circle, if $\overrightarrow{A B}+3 \overrightarrow{B C}+2 \overrightarrow{C D}+4 \overrightarrow{D A}=0$, and $|\overrightarrow{A C}|=4$, then the maximum of $|\overrightarrow{A B}|+|\overrightarrow{B C}|$ is \_\_\_\_\_. | Set $A C$ and $B D$ intersects at point $P$. From the given conditions, \\
$\Leftrightarrow 3 \overrightarrow{P A}+\overrightarrow{P C}=2 \overrightarrow{P B}+2 \overrightarrow{P D}$
$\Rightarrow|\overrightarrow{P A}|:|\overrightarrow{P C}|=1: 3, \ |\overrightarrow{P B}|:|\overrightarrow{P D}|=1: 1$. Furthermore, by the Power of a Point theorem $(|\overrightarrow{P A}|,|\overrightarrow{P C}|,|\overrightarrow{P B}|,|\overrightarrow{P D}|)=(1,3, \sqrt{3}, \sqrt{3})$. Set $\angle A P B=\theta$. From Cauchy's inequality: $|\overrightarrow{A B}|+|\overrightarrow{B C}|=\sqrt{4-2 \sqrt{3} \cos \theta}+\sqrt{3(4+2 \sqrt{3} \cos \theta)}$, $
\sqrt{(1+3)(4+4)}=4 \sqrt{2}$, when $\cos \theta=\frac{\sqrt{3}}{3}$, the equality in the above equation holds. | AIChallenge_Geo7 | 4 \sqrt{2} |
Given that the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is $1$, where, $E$ is the middle point of $A B$, $F$ is the middle point of $C C_{1}$. Then the distance from point $D$ to the plane passing through the three points $D_{1}, E, F$ is \_\_\_\_\_. | Set $D$ as the origin, establish a three-dimensional Cartesian coordinate system respectively with $DA$, $DC$, and $DD_1$ as the $x$, $y$, and $z$ axes, then $D_{1}(0,0,1), E\left(1, \frac{1}{2}, 0\right), F\left(0,1, \frac{1}{2}\right)$.
normal vector $\mathbf{n}=(3,2,4)$ to the plane passing through points $D_{1}, E, F$. Additionally, $\overrightarrow{D D_{1}}=(0,0,1)$, the distance from point $D$ to the plane passing through points $D_{1}, E, F$ is $\frac{\left|\overrightarrow{D D_{1}} \cdot \mathbf{n}\right|}{|\mathbf{n}|}=\frac{4}{29} \sqrt{29}$. | AIChallenge_Geo8 | \frac{4}{29} \sqrt{29} |
Given that the vertices of triangle $\triangle OAB$ are $O(0,0)$, $A(4,4\sqrt{3})$, and $B(8,0)$, with the incenter denoted as $I$, let $\Gamma$ be a circle passing through points $A$ and $B$, intersecting circle $\odot I$ at points $P$ and $Q$. If the tangents drawn through points $P$ and $Q$ are perpendicular, then the radius of circle $\Gamma$ is \_\_\_\_\_. | Let $\odot I$ tangent to $BO$, $AB$, and $AO$ at points $D$, $E$, and $F$, respectively. Due to $\triangle OAB$ being an equilateral triangle, $D(4,0)$, $E(6,2\sqrt{3})$, and $F(2,2\sqrt{3})$.\\
From the given information and the power of a point theorem, we know that circle $\Gamma$ passes through the midpoints of $DE$ and $EF$, denoted as $G(5,\sqrt{3})$ and $H(4,2\sqrt{3})$, respectively.\\
Moreover, since circle $\Gamma$ passes through points $A$ and $B$, the radius of circle $\Gamma$ is found to be $2\sqrt{7}$. | AIChallenge_Geo9 | 2 \sqrt{7} |
A person has some $2 \times 5 \times 8$ bricks and some $2 \times 3 \times 7$ bricks, as well as a $10 \times 11 \times 14$ box. All bricks and the box are rectangular prisms. He wants to pack all the bricks into the box so that the bricks can fill the entire box. The number of bricks he can fit into the box is \_\_\_\_\_ pieces. | Let the number of $2 \times 5 \times 8$ bricks be $a$ and the number of $2 \times 3 \times 7$ bricks be $b$.
According to the problem, we have $2 \times 5 \times 8 \cdot a + 2 \times 3 \times 7 \cdot b = 10 \times 11 \times 14 \Rightarrow 40a + 21b = 770$.
Since $(21,770) = 7$ and $(40,7) = 1$, we know that $7 \mid a$.
So, $40a \equiv 0 \pmod{7} \Rightarrow a \equiv 0 \cdot \frac{77}{4} \equiv 0 \pmod{7}$.
This implies that $a=7$ or $14$.
When $a=7$, $b=\frac{490}{21} = \frac{70}{3} \notin \mathbf{Z}$;
When $a=14$, $b=\frac{210}{21} = 10$, satisfying the requirements. In this case, a total of $14+10=24$ bricks are used.
Next, let's prove that we can indeed use 14 $2 \times 5 \times 8$ bricks and 10 $2 \times 3 \times 7$ bricks to form a $10 \times 11 \times 14$ rectangular prism.
We stack 7 $2 \times 5 \times 8$ bricks vertically to form a $14 \times 5 \times 8$ rectangular prism, and stack the remaining 7 $2 \times 5 \times 8$ bricks vertically to form another $14 \times 5 \times 8$ rectangular prism. Then, we horizontally combine these two $14 \times 5 \times 8$ prisms to form a $14 \times 10 \times 8$ prism.
Similarly, we stack 5 $2 \times 3 \times 7$ bricks vertically to form a $10 \times 3 \times 7$ rectangular prism, and stack the remaining 5 $2 \times 3 \times 7$ bricks vertically to form another $10 \times 3 \times 7$ rectangular prism. Then, we horizontally combine these two $10 \times 3 \times 7$ prisms to form a $10 \times 3 \times 14$ prism.
Finally, we horizontally combine the $10 \times 8 \times 14$ prism with the $10 \times 3 \times 14$ prism, resulting in a $10 \times 11 \times 14$ rectangular prism, meeting the requirements.
Therefore, the required number of bricks is 24. | AIChallenge_Geo10 | 24 |
Let $a$ be an acute angle not exceeding $45^\circ$. If $\cot 2a - \sqrt{3} = \sec a$, then $a =$ \_\_\_\_\_ degrees. | According to the given conditions: $\frac{1}{\cos a}=\cot 2 a-\sqrt{3}=\frac{\frac{1}{2} \cos 2 a-\frac{\sqrt{3}}{2} \sin 2 a}{\frac{1}{2} \sin 2 a}=\frac{\sin \left(30^{\circ}-2 a\right)}{\sin a \cdot \cos a} \Rightarrow \sin a=\sin \left(30^{\circ}-2 a\right)$.
Additionally, $0<a<45^{\circ}$, so $a=10^{\circ}$. | AIChallenge_Geo11 | 10 |
Known that there is a regular 200-gon $A_{1}A_{2} \ldots A_{200}$, connecting the diagonals $A_{i}A_{i+9}(\mathrm{i}=1,2, \ldots, 200)$, where $A_{i+200}=A_{i}(i=1,2, \ldots, 9)$. Then there are a total of \_\_\_\_\_ distinct intersection points inside the regular 200-gon. | Obviously, each diagonal intersects with $8 \times 2 = 16$ other diagonals. Therefore, there are a total of $200 \times 16 \div 2 = 1600$ intersections. Moreover, all these diagonals should be tangent to the same circle, which is concentric with and smaller than the circumscribed circle of the regular 200-gon. Since there can be at most two tangents passing through a point, there are no three lines intersecting at the same point. | AIChallenge_Geo12 | 1600 |
The distance between the highest point of the ellipse obtained by counterclockwise rotating the ellipse $\frac{x^{2}}{2}+y^{2}=1$ about the origin by 45 degrees and the origin is:\_\_\_\_\_. | The tangent line drawn through the highest point should be parallel to the $\mathrm{x}$-axis. Therefore, after clockwise rotation by 45 degrees, this tangent line should have a slope of 1. By utilizing the slope of the tangent line, we can solve for the coordinates of the point of tangency as $\left(\frac{2 \sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$ (discarding the solution in the third quadrant). Consequently, we find the distance from the origin to be $\frac{\sqrt{15}}{3}$. | AIChallenge_Geo13 | \frac{\sqrt{15}}{3} |
A convex $\mathrm{n}$-gon with interior angles of $\mathrm{n}$ degrees each, all integers, and all different.
The degree measure of the largest interior angle is three times the degree measure of the smallest interior angle. The maximum value that $n$ can take is \_\_\_\_\_. | Let the smallest interior angle be $m$ degrees, then $m \leq 59$. The largest interior angle is $3m$ degrees, and the next largest interior angle is at most $3m - 1$ degrees, and so on until the second smallest interior angle (the $(n-1)$-th largest) is at most $3m - n + 2$ degrees. Therefore, we have:
$$
180(n-2) \leq m+3 m+(3 m-1)+(3 m-2)+\ldots+(3 m-n+2)
$$
Using $m \leq 59$ for simplification, we get $n^2 + 3n - 482 \leq 0$. Hence, $n \leq 20$. | AIChallenge_Geo14 | 20 |
In triangle $\mathrm{ABC}$ with its incenter $\mathrm{I}$, if $3\overrightarrow{IA} + 4\overrightarrow{IB} + 5\overrightarrow{IC} = \overrightarrow{0}$, then the measure of angle $\mathrm{C}$ is \_\_\_\_\_. | Extending $I \vec{A}, I \vec{B}$, and $I \vec{C}$ respectively to $3, 4$, and $5$ times, $\mathrm{I}$ becomes the centroid of the resulting new triangle. Then, we can infer that the ratios of the areas of triangles $\mathrm{IAB}, \mathrm{IBC}$, and $\mathrm{ICA}$ are $\frac{1}{3 \times 4}: \frac{1}{4 \times 5}: \frac{1}{5 \times 3}=5: 3: 4$. Furthermore, since $\mathrm{I}$ is the incenter, we have $\mathrm{AB}: \mathrm{BC}: \mathrm{CA}=5: 3: 4$. Finally, by the converse of the Pythagorean theorem, angle $\mathrm{C}$ is a right angle. | AIChallenge_Geo15 | 90 |
Given the circle $x^2 + y^2 = 4$ and the point $\mathrm{P}(2,1)$, two mutually perpendicular lines are drawn through point $\mathrm{P}$, intersecting the circle at points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}, \mathrm{D}$ respectively. Point $\mathrm{A}$ lies inside the line segment $PB$, and point $\mathrm{D}$ lies inside the line segment $PC$. The maximum area of quadrilateral $ABCD$ is \_\_\_\_\_. | Set midpoints of $A B, C D$ to be $M, N$ relatively.
$$
S_{A B C D}=\frac{1}{2}(P B \cdot P C-P A \cdot P D)=\frac{1}{2}[(P M+M B)(P N+N C)-(P M-
$$
$M A)(P N-N D)]=\mathrm{PM} \cdot \mathrm{NC}+\mathrm{PN} \cdot \mathrm{MB}$
By the Cauchy inequality
$$
\begin{aligned}
& (\mathrm{PM} \cdot \mathrm{NC}+\mathrm{PN} \cdot \mathrm{MB})^{2} \leq\left(P M^{2}+P N^{2}\right)\left(M B^{2}+M C^{2}\right) \\
& =O P^{2}\left(2^{2}-O M^{2}+2^{2}-O N^{2}\right)=O P^{2}\left(8-O P^{2}\right)=15
\end{aligned}
$$
Hence, the maximum value of the quadrilateral $\mathrm{ABCD}$ area is $\sqrt{15}$. | AIChallenge_Geo16 | \sqrt{15} |